# 4.E: Exercises

- Page ID
- 93323

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Find \(-3\left[\begin{array}{c}5\\-1\\2\\-3\end{array}\right]+5\left[\begin{array}{c}-8\\2\\-3\\6\end{array}\right]\).

**Answer**-
\(\left[\begin{array}{c}-55\\13\\-21\\39\end{array}\right]\)

Find \(-7\left[\begin{array}{c}6\\0\\4\\-1\end{array}\right]+6\left[\begin{array}{c}-13\\-1\\1\\6\end{array}\right]\).

Decide whether \[\vec{v}=\left[\begin{array}{c}4\\4\\-3\end{array}\right]\nonumber\] is a linear combination of the vectors \[\vec{u}_{1}=\left[\begin{array}{c}3\\1\\-1\end{array}\right]\quad\text{and}\quad\vec{u}_{2}=\left[\begin{array}{c}2\\-2\\1\end{array}\right].\nonumber\]

**Answer**-
\[\left[\begin{array}{c}4\\4\\-3\end{array}\right]=2\left[\begin{array}{c}3\\1\\-1\end{array}\right]-\left[\begin{array}{c}2\\-2\\1\end{array}\right]\nonumber\]

Decide whether \[\vec{v}=\left[\begin{array}{c}4\\4\\4\end{array}\right]\nonumber\] is a linear combination of the vectors \[\vec{u}_1=\left[\begin{array}{c}3\\1\\-1\end{array}\right]\quad\text{and}\quad\vec{u}_2=\left[\begin{array}{c}2\\-2\\1\end{array}\right].\nonumber\]

**Answer**-
The system \[\left[\begin{array}{c}4\\4\\4\end{array}\right]=a_1\left[\begin{array}{c}3\\1\\-1\end{array}\right]+a_2\left[\begin{array}{c}2\\-2\\1\end{array}\right]\nonumber\] has no solution.

Find the vector equation for the line through \((−7, 6, 0)\) and \((−1, 1, 4)\). Then, find the parametric equations for this line.

Find parametric equations for the line through the point \((7, 7, 1)\) with a direction vector \(\vec{d}=\left[\begin{array}{c}1\\6\\2\end{array}\right]\).

Parametric equations of the line are \[\begin{aligned}x&=t+2 \\ y&=6-3t \\ x&=-t=6\end{aligned}\] Find a direction vector for the line and a point on the line.

Find the vector equation for the line through the two points \((−5, 5, 1),\: (2, 2, 4)\). Then, find the parametric equations.

The equation of a line in two dimensions is written as \(y = x−5\). Find parametric equations for this line.

Find parametric equations for the line through \((6, 5,−2)\) and \((5, 1, 2)\).

Find the vector equation and parametric equations for the line through the point \((−7, 10,−6)\) with a direction vector \(\vec{d}=\left[\begin{array}{c}1\\1\\3\end{array}\right]\).

Parametric equations of the line are \[\begin{aligned}x&=2t+2 \\ y&=5-4t \\ z&=-t-3\end{aligned}\] Find a direction vector for the line and a point on the line, and write the vector equation of the line.

Find the vector equation and parametric equations for the line through the two points \((4, 10, 0),\: (1,−5,−6)\).

Find the point on the line segment from \(P = (−4, 7, 5)\) to \(Q = (2,−2,−3)\) which is \(\frac{1}{7}\) of the way from \(P\) to \(Q\).

Suppose a triangle in \(\mathbb{R}^n\) has vertices at \(P_1,\: P_2,\) and \(P_3\). Consider the lines which are drawn from a vertex to the mid point of the opposite side. Show these three lines intersect in a point and find the coordinates of this point.

Find \(\left[\begin{array}{c}1\\2\\3\\4\end{array}\right]\bullet\left[\begin{array}{c}2\\0\\1\\3\end{array}\right]\).

**Answer**-
\(\left[\begin{array}{c}1\\2\\3\\4\end{array}\right]\bullet\left[\begin{array}{c}2\\0\\1\\3\end{array}\right]=17\)

Use the formula given in Proposition 4.7.2 to verify the Cauchy Schwarz inequality and to show that equality occurs if and only if one of the vectors is a scalar multiple of the other

**Answer**-
This formula says that \(\vec{u}\bullet\vec{v} = ||\vec{u}||\:||\vec{v}||\cos\theta\) where \(θ\) is the included angle between the two vectors. Thus \[||\vec{u}\bullet\vec{v}||=||\vec{u}||\:||\vec{v}||\:||\cos\theta||\leq ||\vec{u}||\:||\vec{v}||\nonumber\] and equality holds if and only if \(\theta = 0\) or \(π\). This means that the two vectors either point in the same direction or opposite directions. Hence one is a multiple of the other.

For \(\vec{u}\), \(\vec{v}\) vectors in \(\mathbb{R}^3\), define the product, \(\vec{u}\ast\vec{v} = u_1v_1 +2u_2v_2 +3u_3v_3\). Show the axioms for a dot product all hold for this product. Prove \[||\vec{u}\ast\vec{v}||\leq (\vec{u}\ast\vec{u})^{1/2}(\vec{v}\ast\vec{v})^{1/2}\nonumber\]

**Answer**-
This follows from the Cauchy Schwarz inequality and the proof of Theorem 4.7.1 which only used the properties of the dot product. Since this new product has the same properties the Cauchy Schwarz inequality holds for it as well.

Let \(\vec{a}\), \(\vec{b}\) be vectors. Show that \(\left(\vec{a}\bullet\vec{b}\right)=\frac{1}{4}\left(||\vec{a}+\vec{b}||^2-||\vec{a}-\vec{b}||^2\right).\)

Using the axioms of the dot product, prove the parallelogram identity: \[||\vec{a}+\vec{b}||^2+||\vec{a}-\vec{b}||^2=2||\vec{a}||^2+2||\vec{b}||^2\nonumber\]

Let \(A\) be a real \(m\times n\) matrix and let \(\vec{u} ∈ \mathbb{R}^n\) and \(\vec{v} ∈ \mathbb{R}^m\). Show \(A\vec{u}\bullet\vec{v} =\vec{u}\bullet A^T\vec{v}\). Hint: Use the definition of matrix multiplication to do this.

**Answer**-
\(A\vec{x}\bullet\vec{y}=\sum_k(A\vec{x})_ky_k=\sum_k\sum_iA_{ki}x_iy_k=\sum_i\sum_kA^T_{ik}x_iy_k=\vec{x}\bullet A^T\vec{y}\)

Use the result of Problem \(\PageIndex{21}\) to verify directly that \((AB)^T = B^TA^T\) without making any reference to subscripts.

**Answer**-
\[\begin{aligned}AB\vec{x}\bullet\vec{y}&=B\vec{x}\bullet A^T\vec{y} \\ &=\vec{x}\bullet B^TA^T\vec{y} \\ &=\vec{x}\bullet (AB)^T\vec{y}\end{aligned}\] Since this is true for all \(\vec{x}\), it follows that, in particular, it holds for \[\vec{x}=B^TA^T\vec{y}-(AB)^T\vec{y}\nonumber\] and so from the axioms of the dot product, \[\left(B^TA^T\vec{y}-(AB)^T\vec{y}\right)\bullet\left(B^TA^T\vec{y}-(AB)^T\vec{y}\right)=0\nonumber\] and so \(B^TA^T\vec{y}-(AB)^T\vec{y}=\vec{0}\). However, this is true for all \(\vec{y}\) and so \(B^TA^T-(AB)^T=0\).

Find the angle between the vectors \[\vec{u}=\left[\begin{array}{r}3\\-1\\-1\end{array}\right],\:\vec{v}=\left[\begin{array}{c}1\\4\\2\end{array}\right]\nonumber\]

**Answer**-
\(\frac{\left[\begin{array}{ccc}3&-1&-1\end{array}\right]^T\bullet\left[\begin{array}{ccc}1&4&2\end{array}\right]^T}{\sqrt{9+1+1}\sqrt{1+16+4}}=\frac{-3}{\sqrt{11}\sqrt{21}}=-0.19739=\cos\theta\) Therefore we need to solve \[-0.19739=\cos\theta\nonumber\] Thus \(\theta=1.7695\) radians.

Find the angle between the vectors \[\vec{u}=\left[\begin{array}{r}1\\-2\\1\end{array}\right],\:\vec{v}=\left[\begin{array}{r}1\\2\\-7\end{array}\right]\nonumber\]

**Answer**-
\(\frac{-10}{\sqrt{1+4+1}\sqrt{1+4+49}}=-0.55555=\cos\theta\) Therefore we need to solve \(−0.55555 = \cos θ\), which gives \(θ = 2.0313\) radians.

Find \(\text{proj}_{\vec{v}}(\vec{w})\) where \(\vec{w}=\left[\begin{array}{r}1\\0\\-2\end{array}\right]\) and \(\vec{v}=\left[\begin{array}{c}1\\2\\3\end{array}\right]\).

**Answer**-
\(\frac{\vec{u}\bullet\vec{v}}{\vec{u}\bullet\vec{u}}\vec{u}=\frac{-5}{14}\left[\begin{array}{c}1\\2\\3\end{array}\right]=\left[\begin{array}{r}-\frac{5}{14}\\-\frac{5}{7}\\-\frac{15}{14}\end{array}\right]\)

Find \(\text{proj}_{\vec{v}}(\vec{w})\) where \(\vec{w}=\left[\begin{array}{r}1\\2\\-2\end{array}\right]\) and \(\vec{v}=\left[\begin{array}{c}1\\0\\3\end{array}\right]\).

**Answer**-
\(\frac{\vec{u}\bullet\vec{v}}{\vec{u}\bullet\vec{u}}\vec{u}=\frac{-5}{10}\left[\begin{array}{c}1\\0\\3\end{array}\right]=\left[\begin{array}{r}-\frac{1}{2}\\0\\-\frac{3}{2}\end{array}\right]\)

Find \(\text{proj}_{\vec{v}}(\vec{w})\) where \(\vec{w}=\left[\begin{array}{r}1\\2\\-2\\1\end{array}\right]\) and \(\vec{v}=\left[\begin{array}{c}1\\2\\3\\0\end{array}\right]\).

**Answer**-
\(\frac{\vec{u}\bullet\vec{v}}{\vec{u}\bullet\vec{u}}\vec{u}=\frac{\left[\begin{array}{cccc}1&2&-2&1\end{array}\right]^T\bullet\left[\begin{array}{cccc}1&2&3&0\end{array}\right]^T}{1+4+9}\left[\begin{array}{c}1\\2\\3\\0\end{array}\right]=\left[\begin{array}{r}-\frac{1}{14}\\-\frac{1}{7}\\-\frac{3}{14}\\0\end{array}\right]\)

Let \(P = (1, 2, 3)\) be a point in \(\mathbb{R}^3\). Let \(L\) be the line through the point \(P_0 = (1, 4, 5)\) with direction vector \(\vec{d} =\left[\begin{array}{r}1\\-1\\1\end{array}\right]\). Find the shortest distance from \(P\) to \(L\), and find the point \(Q\) on \(L\) that is closest to \(P\).

Let \(P = (0, 2, 1)\) be a point in \(\mathbb{R}^3\). Let \(L\) be the line through the point \(P_0 = (1, 1, 1)\) with direction vector \(\vec{d} =\left[\begin{array}{c}3\\0\\1\end{array}\right]\). Find the shortest distance from \(P\) to \(L\), and find the point \(Q\) on \(L\) that is closest to \(P\).

Does it make sense to speak of \(\text{proj}_{\vec{0}} (\vec{w})\)?

**Answer**-
No, it does not. The \(0\) vector has no direction. The formula for \(\text{proj}_{\vec{0}} (\vec{w})\) doesn’t make sense either.

Prove the Cauchy Schwarz inequality in \(\mathbb{R}^n\) as follows. For \(\vec{u}\),\(\vec{v}\) vectors, consider \[(\vec{w}-\text{proj}_{\vec{v}}\vec{w})\bullet (\vec{w}-\text{proj}_{\vec{v}}\vec{w})\geq 0\nonumber\] Simplify using the axioms of the dot product and then put in the formula for the projection. Notice that this expression equals \(0\) and you get equality in the Cauchy Schwarz inequality if and only if \(\vec{w} = \text{proj}_{\vec{v}}\vec{w}\). What is the geometric meaning of \(\vec{w}= \text{proj}_{\vec{v}}\vec{w}\)?

**Answer**-
\[\left(\vec{u}-\frac{\vec{u}\bullet\vec{v}}{||\vec{v}||^2}\vec{v}\right)\bullet\left(\vec{u}-\frac{\vec{u}\bullet\vec{v}}{||\vec{v}||^2}\vec{v}\right)=||\vec{u}||^2-2(\vec{u}\bullet\vec{v})^2\frac{1}{||\vec{v}||^2}+(\vec{u}\bullet\vec{v})^2\frac{1}{||\vec{v}||^2}\geq 0\nonumber\] And so \[||\vec{u}||^2||\vec{v}||^2\geq (\vec{u}\bullet\vec{v})^2\nonumber\] You get equality exactly when \(\vec{u}=\text{proj}_{\vec{v}}\vec{u}=\frac{\vec{u}\bullet\vec{v}}{||\vec{v}||^2}\vec{v}\) in other words, when \(\vec{u}\) is a multiple of \(\vec{v}\).

Let \(\vec{v},\:\vec{w},\:\vec{u}\) be vectors. Show that \((\vec{w}+\vec{u})_{\perp}=\vec{w}_\perp +\vec{u}_\perp\) where \(\vec{w}_\perp =\vec{w}-\text{proj}_{\vec{v}}(\vec{w})\).

**Answer**-
\[\begin{aligned}\vec{w}-\text{proj}_{\vec{v}}(\vec{w})+\vec{u}-\text{proj}_{\vec{v}}(\vec{u})&=\vec{w}+\vec{u}-(\text{proj}_{\vec{v}}(\vec{w})+\text{proj}_{\vec{v}}(\vec{u})) \\ &=\vec{w}+\vec{u}-\text{proj}_{\vec{v}}(\vec{w}+\vec{u})\end{aligned}\] This follows because \[\begin{aligned}\text{proj}_{\vec{v}}(\vec{w})+\text{proj}_{\vec{v}}(\vec{u})&=\frac{\vec{u}\bullet\vec{v}}{||\vec{v}||^2}\vec{v}+\frac{\vec{w}\bullet\vec{v}}{||\vec{v}||^2}\vec{v} \\ &=\frac{(\vec{u}+\vec{w})\bullet\vec{v}}{||\vec{v}||^2}\vec{v} \\ &=\text{proj}_{\vec{v}}(\vec{w}+\vec{u})\end{aligned}\]

Show that \[(\vec{v}-\text{proj}_{\vec{u}}(\vec{v}),\vec{u})=(\vec{v}-\text{proj}_{\vec{u}}(\vec{v}))\bullet\vec{u}=0\nonumber\] and conclude every vector in \(\mathbb{R}^n\) can be written as the sum of two vectors, one which is perpendicular and one which is parallel to the given vector.

**Answer**-
\((\vec{v}-\text{proj}_{\vec{u}}(\vec{v}))\bullet\vec{u}=\vec{v}\bullet\vec{u}-\left(\frac{(\vec{v}\cdot\vec{u}}{||\vec{u}||^2}\vec{u}\right)\bullet\vec{u}=\vec{v}\bullet\vec{u}-\vec{v}\bullet\vec{u}=0\). Therefore, \(\vec{v}=\vec{v}-\text{proj}_{\vec{u}}(\vec{v})+\text{proj}_{\vec{u}}(\vec{v})\). The first is perpendicular to \(\vec{u}\) and the second is a multiple of \(\vec{u}\) so it is parallel to \(\vec{u}\).

Show that if \(\vec{a}\times\vec{u}=\vec{0}\) for any unit vector \(\vec{u}\), then \(\vec{a}=\vec{0}\).

**Answer**-
If \(\vec{a}\neq\vec{0}\), then the condition says that \(||\vec{a}\times\vec{u}||=||\vec{a}||\sin\theta =0\) for all angles \(θ\). Hence \(\vec{a}=\vec{0}\) after all.

Find the area of the triangle determined by the three points \((1, 2, 3),\: (4, 2, 0)\) and \((−3, 2, 1)\).

**Answer**-
\(\left[\begin{array}{r}3\\0\\-3\end{array}\right]\times\left[\begin{array}{r}-4\\0\\-2\end{array}\right]=\left[\begin{array}{r}0\\18\\0\end{array}\right]\). So the area is \(9\).

Find the area of the triangle determined by the three points \((1, 0, 3),\: (4, 1, 0)\) and \((−3, 1, 1)\).

**Answer**-
\(\left[\begin{array}{r}3\\1\\-3\end{array}\right]\times\left[\begin{array}{r}-4\\1\\-2\end{array}\right]=\left[\begin{array}{c}1\\18\\7\end{array}\right]\). The area is given by \[\frac{1}{2}\sqrt{1+(18)^2+49}=\frac{1}{2}\sqrt{374}\nonumber\]

Find the area of the triangle determined by the three points, \((1, 2, 3),\: (2, 3, 4)\) and \((3, 4, 5)\). Did something interesting happen here? What does it mean geometrically?

**Answer**-
\(\left[\begin{array}{ccc}1&1&1\end{array}\right]\times\left[\begin{array}{ccc}2&2&2\end{array}\right]=\left[\begin{array}{ccc}0&0&0\end{array}\right]\). The area is \(0\). It means the three points are on the same line.

Find the area of the parallelogram determined by the vectors \(\left[\begin{array}{c}1\\2\\3\end{array}\right]\), \(\left[\begin{array}{r}3\\-2\\1\end{array}\right]\).

**Answer**-
\(\left[\begin{array}{c}1\\2\\3\end{array}\right]\times\left[\begin{array}{r}3\\-2\\1\end{array}\right]=\left[\begin{array}{r}8\\8\\-8\end{array}\right]\). The area is \(8\sqrt{3}\).

Find the area of the parallelogram determined by the vectors \(\left[\begin{array}{c}1\\0\\3\end{array}\right]\), \(\left[\begin{array}{r}4\\-2\\1\end{array}\right]\).

**Answer**-
\(\left[\begin{array}{c}1\\0\\3\end{array}\right]\times\left[\begin{array}{r}4\\-2\\1\end{array}\right]=\left[\begin{array}{r}6\\11\\-2\end{array}\right]\). The area is \(\sqrt{36+121+4}=\sqrt{161}\).

Is \(\vec{u}\times (\vec{v}\times\vec{w})=(\vec{u}\times\vec{v})\times\vec{w}\)? What is the meaning of \(\vec{u}\times\vec{v}\times\vec{w}\)? Explain. Hint: Try \(\left(\vec{i}\times\vec{j}\right)\times\vec{k}\).

**Answer**-
\(\left(\vec{i}\times\vec{j}\right)\times\vec{j}=\vec{k}\times\vec{j}=i\vec{i}\). However, \(\vec{i}\times\left(\vec{j}\times\vec{j}\right)=\vec{0}\) and so the cross product is not associative.

Verify directly that the coordinate description of the cross product, \(\vec{u}\times\vec{v}\) has the property that it is perpendicular to both \(\vec{u}\) and \(\vec{v}\). Then show by direct computation that this coordinate description satisfies \[\begin{aligned} ||\vec{u}\times\vec{v}||^2&=||\vec{u}||^2||\vec{v}||^2-(\vec{u}\bullet\vec{v})^2 \\ &=||\vec{u}||^2||\vec{v}||^2(1-\cos^2(\theta ))\end{aligned}\] where \(\theta\) is the angle included between the two vectors. Explain why \(||\vec{u}\times\vec{v}||\) has the correct magnitude.

**Answer**-
Verify directly from the coordinate description of the cross product that the right hand rule applies to the vectors \(\vec{i},\vec{j},\vec{k}\). Next verify that the distributive law holds for the coordinate description of the cross product. This gives another way to approach the cross product. First define it in terms of coordinates and then get the geometric properties from this. However, this approach does not yield the right hand rule property very easily. From the coordinate description, \[\vec{a}\times\vec{b}\cdot\vec{a}=\epsilon_{ijk}a_jb_ka_i=-\epsilon_{jik}a_kb_ka_i=-\epsilon_{jik}b_ka_ia_j=-\vec{a}\times\vec{b}\cdot\vec{a}\nonumber\] and so \(\vec{a}\times\vec{b}\) is perpendicular to \(\vec{a}\). Similarly, \(\vec{a}\times\vec{b}\) is perpendicular to \(\vec{b}\). Now we need that \[||\vec{a}\times\vec{b}||^2=||\vec{a}||^2||\vec{b}||^2(1-\cos^2\theta )=||\vec{a}||^2||\vec{b}||^2\sin^2\theta\nonumber\] and so \(||\vec{a}\times\vec{b}||=||\vec{a}||\:||\vec{b}||\sin\theta\), the area of the parallelogram determined by \(\vec{a}\), \(\vec{b}\). Only the right hand rule is a little problematic. However, you can see right away from the component definition that the right hand rule holds for each of the standard unit vectors. Thus \(\vec{i}\times\vec{j}=\vec{k}\) etc. \[\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\1&0&0\\0&1&0\end{array}\right|=\vec{k}\nonumber\]

Suppose \(A\) is a \(3\times 3\) skew symmetric matrix such that \(A^T = −A\). Show there exists a vector \(\vec{Ω}\) such that for all \(\vec{u} ∈ \mathbb{R}^3\) \[A\vec{u}=\vec{\Omega}\times\vec{u}\nonumber\] Hint: Explain why since \(A\) is skew symmetric it is of the form \[A=\left[\begin{array}{ccc}0&-\omega_3&\omega_2 \\ \omega_3&0&-\omega_1 \\ -\omega_2&\omega_1&0\end{array}\right]\nonumber\] where the \(\omega_i\) are numbers. Then consider \(\omega_1\vec{i}+\omega_2\vec{j}+\omega_3\vec{k}\).

Find the volume of the parallelepiped determined by the vectors \(\left[\begin{array}{r}1\\-7\\-5\end{array}\right]\), \(\left[\begin{array}{r}1\\-2\\-6\end{array}\right]\), and \(\left[\begin{array}{c}3\\2\\3\end{array}\right]\).

**Answer**-
\(\left|\begin{array}{ccc}1&-7&-5 \\ 1&-2&-6 \\ 3&2&3\end{array}\right|=113\)

Suppose \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\) are three vectors whose components are all integers. Can you conclude the volume of the parallelepiped determined from these three vectors will always be an integer?

**Answer**-
Yes. It will involve the sum of product of integers and so it will be an integer.

What does it mean geometrically if the box product of three vectors gives zero?

**Answer**-
It means that if you place them so that they all have their tails at the same point, the three will lie in the same plane.

Using Problem \(\PageIndex{45}\), find an equation of a plane containing the two position vectors, \(\vec{p}\) and \(\vec{q}\) and the point \(0\). Hint: If \((x, y,z)\) is a point on this plane, the volume of the parallelepiped determined by \((x, y,z)\) and the vectors \(\vec{p}\), \(\vec{q}\) equals \(0\).

**Answer**-
\(\vec{x}\bullet\left(\vec{a}\times\vec{b}\right)=0\)

Using the notion of the box product yielding either plus or minus the volume of the parallelepiped determined by the given three vectors, show that \[(\vec{u}\times\vec{v})\bullet\vec{w}=\vec{u}\bullet (\vec{v}\times\vec{w})\nonumber\] In other words, the dot and the cross can be switched as long as the order of the vectors remains the same. Hint: There are two ways to do this, by the coordinate description of the dot and cross product and by geometric reasoning.

Simplify \((\vec{u}\times\vec{v})\bullet (\vec{v}\times\vec{w})\times (\vec{w}\times\vec{z})\).

**Answer**-
Here \([\vec{v},\vec{w},\vec{z}]\) denotes the box product. Consider the cross product term. From the above, \[\begin{aligned}(\vec{v}\times\vec{w})\times(\vec{w}\times\vec{z})&=[\vec{v},\vec{w},\vec{z}]\vec{w}-[\vec{w},\vec{w},\vec{z}]\vec{v} \\ &=[\vec{v},\vec{w},\vec{z}]\vec{w}\end{aligned}\] Thus it reduces to \[(\vec{u}\times\vec{v})\bullet [\vec{v},\vec{w},\vec{z}]\vec{w}=[\vec{v},\vec{w},\vec{z}][\vec{u},\vec{v},\vec{w}]\nonumber\]

Simplify \(||\vec{u}\times\vec{v}||^2+(\vec{u}\bullet\vec{v})^2-||\vec{u}||^2||\vec{v}||^2\).

**Answer**-
\[\begin{aligned}||\vec{u}\times\vec{v}||^2&=\epsilon_{ijk}u_jv_k\epsilon_{irs}u_rv_s=(\delta_{jr}\delta_{ks}-\delta_{kr}\delta_{js})u_rv_su_jv_k \\ &=u_jv_ku_jv_k-u_kv_ju_jv_k=||\vec{u}||^2||\vec{v}||^2-(\vec{u}\bullet\vec{v})^2\end{aligned}\] It follows that the expression reduces to \(0\). You can also do the following. \[\begin{aligned}||\vec{u}\times\vec{v}||^2&=||\vec{u}||^2||\vec{v}||^2\sin^2\theta \\ &=||\vec{u}||^2||\vec{v}||^2(1-\cos^2\theta ) \\ &=||\vec{u}||^2||\vec{v}||^2-||\vec{u}||^2||\vec{v}||^2\cos^2\theta \\ &=||\vec{u}||^2||\vec{v}||^2-(\vec{u}\bullet\vec{v})^2\end{aligned}\] which implies the expression equals \(0\).

For \(\vec{u},\:\vec{v},\:\vec{w}\) functions of \(t\), prove the following product rules: \[\begin{aligned}(\vec{u}\times\vec{v})'&=\vec{u}'\times\vec{v}+\vec{u}\times\vec{v}' \\ (\vec{u}\bullet\vec{v})'&=\vec{u}'\bullet\vec{v}+\vec{u}\bullet\vec{v}'\end{aligned}\]

**Answer**-
We will show it using the summation convention and permutation symbol \[\begin{aligned}((\vec{u}\times\vec{v})')_i&=((\vec{u}\times\vec{v})_i)'=(\epsilon_{ijk}u_jv_k)' \\ &=\epsilon_{ijk}u_j'v_k+\epsilon_{ijk}u_kv_k'=(\vec{u}'\times\vec{v}+\vec{u}\times\vec{v}')_i\end{aligned}\] and so \((\vec{u}\times\vec{v})'=\vec{u}'\times\vec{v}+\vec{u}\times\vec{v}'\).

Here are some vectors. \[\left[\begin{array}{r}1\\1\\-2\end{array}\right],\:\left[\begin{array}{r}1\\2\\-2\end{array}\right],\:\left[\begin{array}{r}2\\7\\-4\end{array}\right],\:\left[\begin{array}{r}5\\7\\-10\end{array}\right],\:\left[\begin{array}{r}12\\17\\-24\end{array}\right]\nonumber\] Describe the span of these vectors as the span of as few vectors as possible.

Here are some vectors. \[\left[\begin{array}{r}1\\2\\-2\end{array}\right],\:\left[\begin{array}{r}12\\29\\-24\end{array}\right],\:\left[\begin{array}{r}1\\3\\-2\end{array}\right],\:\left[\begin{array}{r}2\\9\\-4\end{array}\right],\:\left[\begin{array}{r}5\\12\\-10\end{array}\right].\nonumber\] Describe the span of these vectors as the span of as few vectors as possible.

Here are some vectors. \[\left[\begin{array}{r}1\\2\\-2\end{array}\right],\:\left[\begin{array}{r}1\\3\\-2\end{array}\right],\:\left[\begin{array}{r}1\\-2\\-2\end{array}\right],\:\left[\begin{array}{r}-1\\0\\2\end{array}\right],\:\left[\begin{array}{r}1\\3\\-1\end{array}\right]\nonumber\] Describe the span of these vectors as the span of as few vectors as possible.

Here are some vectors. \[\left[\begin{array}{r}1\\1\\-2\end{array}\right],\:\left[\begin{array}{r}1\\2\\-2\end{array}\right],\:\left[\begin{array}{r}1\\-3\\-2\end{array}\right],\:\left[\begin{array}{r}-1\\1\\2\end{array}\right]\nonumber\] Now here is another vector: \[\left[\begin{array}{r}1\\2\\-1\end{array}\right]\nonumber\] Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Here are some vectors. \[\left[\begin{array}{r}1\\1\\-2\end{array}\right],\:\left[\begin{array}{r}1\\2\\-2\end{array}\right],\:\left[\begin{array}{r}1\\-3\\-2\end{array}\right],\:\left[\begin{array}{r}-1\\1\\2\end{array}\right]\nonumber\] Now here is another vector: \[\left[\begin{array}{r}2\\-3\\-4\end{array}\right]\nonumber\] Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Here are some vectors. \[\left[\begin{array}{r}1\\1\\-2\end{array}\right],\:\left[\begin{array}{r}1\\2\\-2\end{array}\right],\:\left[\begin{array}{r}1\\-3\\-2\end{array}\right],\:\left[\begin{array}{r}1\\2\\-1\end{array}\right]\nonumber\] Now here is another vector: \[\left[\begin{array}{r}1\\9\\1\end{array}\right]\nonumber\] Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Here are some vectors, \[\left[\begin{array}{r}1\\-1\\-2\end{array}\right],\:\left[\begin{array}{r}1\\0\\-2\end{array}\right],\:\left[\begin{array}{r}1\\-5\\-2\end{array}\right],\:\left[\begin{array}{r}-1\\5\\2\end{array}\right]\nonumber\] Now here is another vector: \[\left[\begin{array}{r}1\\1\\-1\end{array}\right]\nonumber\] Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Here are some vectors. \[\left[\begin{array}{r}1\\-1\\-2\end{array}\right],\:\left[\begin{array}{r}1\\0\\-2\end{array}\right],\:\left[\begin{array}{r}1\\-5\\-2\end{array}\right],\:\left[\begin{array}{r}-1\\5\\2\end{array}\right]\nonumber\] Now here is another vector: \[\left[\begin{array}{r}1\\1\\-1\end{array}\right]\nonumber\] Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Here are some vectors. \[\left[\begin{array}{r}1\\0\\-2\end{array}\right],\:\left[\begin{array}{r}1\\1\\-2\end{array}\right],\:\left[\begin{array}{r}2\\-2\\-3\end{array}\right],\:\left[\begin{array}{r}-1\\4\\2\end{array}\right]\nonumber\] Now here is another vector: \[\left[\begin{array}{r}-1\\-4\\2\end{array}\right]\nonumber\] Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Suppose \(\{\vec{x}_1,\cdots ,\vec{x}_k\}\) is a set of vectors from \(\mathbb{R}^n\). Show that \(\vec{0}\) is in \(span\{\vec{x}_1,\cdots ,\vec{x}_k\}\).

**Answer**-
\(\sum\limits_{i=1}^k 0\vec{x}_k=\vec{0}\)

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\3\\-1\\1\end{array}\right],\:\left[\begin{array}{r}1\\4\\-1\\1\end{array}\right],\:\left[\begin{array}{r}1\\4\\0\\1\end{array}\right],\:\left[\begin{array}{r}1\\10\\2\\1\end{array}\right]\nonumber\]

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}-1\\-2\\2\\3\end{array}\right],\:\left[\begin{array}{r}-3\\-4\\3\\3\end{array}\right],\:\left[\begin{array}{r}0\\-1\\4\\3\end{array}\right],\:\left[\begin{array}{r}0\\-1\\6\\4\end{array}\right]\nonumber\]

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\5\\-2\\1\end{array}\right],\:\left[\begin{array}{r}1\\6\\-3\\1\end{array}\right],\:\left[\begin{array}{r}-1\\-4\\1\\-1\end{array}\right],\:\left[\begin{array}{r}1\\6\\-2\\1\end{array}\right]\nonumber\]

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\-1\\3\\1\end{array}\right],\:\left[\begin{array}{r}1\\6\\34\\1\end{array}\right],\:\left[\begin{array}{r}1\\0\\7\\1\end{array}\right],\:\left[\begin{array}{r}1\\0\\8\\1\end{array}\right]\nonumber\]

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. \[\left[\begin{array}{r}1\\3\\-1\\1\end{array}\right],\:\left[\begin{array}{r}1\\4\\-1\\1\end{array}\right],\:\left[\begin{array}{r}-3\\-10\\3\\-3\end{array}\right],\:\left[\begin{array}{r}1\\4\\0\\1\end{array}\right]\nonumber\]

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\3\\-3\\1\end{array}\right],\:\left[\begin{array}{r}1\\4\\-5\\1\end{array}\right],\:\left[\begin{array}{r}1\\4\\-4\\1\end{array}\right],\:\left[\begin{array}{r}1\\10\\-14\\1\end{array}\right]\nonumber\]

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\0\\3\\1\end{array}\right],\:\left[\begin{array}{r}1\\1\\8\\1\end{array}\right],\:\left[\begin{array}{r}1\\7\\34\\1\end{array}\right],\:\left[\begin{array}{r}1\\1\\7\\1\end{array}\right]\nonumber\]

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}1\\4\\-2\\1\end{array}\right],\:\left[\begin{array}{r}1\\5\\-3\\1\end{array}\right],\:\left[\begin{array}{r}1\\7\\-5\\1\end{array}\right],\:\left[\begin{array}{r}1\\5\\-2\\1\end{array}\right]\nonumber\]

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. \[\left[\begin{array}{r}1\\2\\2\\-4\end{array}\right],\:\left[\begin{array}{r}3\\4\\1\\-4\end{array}\right],\:\left[\begin{array}{r}0\\-1\\0\\4\end{array}\right],\:\left[\begin{array}{r}0\\-1\\-2\\5\end{array}\right]\nonumber\]

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \[\left[\begin{array}{r}2\\3\\1\\-3\end{array}\right],\:\left[\begin{array}{r}-5\\-6\\0\\3\end{array}\right],\:\left[\begin{array}{r}-1\\-2\\1\\3\end{array}\right],\:\left[\begin{array}{r}-1\\-2\\0\\4\end{array}\right]\nonumber\]

Here are some vectors in \(\mathbb{R}^4\). \[\left[\begin{array}{r}1\\1\\-1\\1\end{array}\right],\:\left[\begin{array}{r}1\\2\\-1\\1\end{array}\right],\:\left[\begin{array}{r}1\\-2\\-1\\1\end{array}\right],\:\left[\begin{array}{r}1\\2\\0\\1\end{array}\right],\:\left[\begin{array}{r}1\\-1\\-1\\1\end{array}\right]\nonumber\] These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Here are some vectors in \(\mathbb{R}^4\). \[\left[\begin{array}{r}1\\2\\-2\\1\end{array}\right],\:\left[\begin{array}{r}1\\3\\-3\\1\end{array}\right],\:\left[\begin{array}{r}1\\3\\-2\\1\end{array}\right],\:\left[\begin{array}{r}4\\3\\-1\\4\end{array}\right],\:\left[\begin{array}{r}1\\3\\-2\\1\end{array}\right]\nonumber\] These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Here are some vectors in \(\mathbb{R}^4\). \[\left[\begin{array}{r}1\\1\\0\\1\end{array}\right],\:\left[\begin{array}{r}1\\2\\1\\1\end{array}\right],\:\left[\begin{array}{r}1\\-2\\-3\\1\end{array}\right],\:\left[\begin{array}{r}2\\-5\\-7\\2\end{array}\right],\:\left[\begin{array}{r}1\\2\\2\\1\end{array}\right]\nonumber\] These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Here are some vectors in \(\mathbb{R}^4\). \[\left[\begin{array}{r}1\\2\\-2\\1\end{array}\right],\:\left[\begin{array}{r}1\\3\\-3\\1\end{array}\right],\:\left[\begin{array}{r}1\\-1\\1\\1\end{array}\right],\:\left[\begin{array}{r}2\\-3\\3\\2\end{array}\right],\:\left[\begin{array}{r}1\\3\\-2\\1\end{array}\right]\nonumber\] These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Here are some vectors in \(\mathbb{R}^4\). \[\left[\begin{array}{r}1\\4\\-2\\1\end{array}\right],\:\left[\begin{array}{r}1\\5\\-3\\1\end{array}\right],\:\left[\begin{array}{r}1\\5\\-2\\1\end{array}\right],\:\left[\begin{array}{r}4\\11\\-1\\4\end{array}\right],\:\left[\begin{array}{r}1\\5\\-3\\1\end{array}\right]\nonumber\] These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Here are some vectors in \(\mathbb{R}^4\). \[\left[\begin{array}{r}1\\3\\-1\\1\end{array}\right],\:\left[\begin{array}{r}-\frac{3}{2}\\-\frac{9}{2}\\ \frac{3}{2}\\ -\frac{3}{2}\end{array}\right],\:\left[\begin{array}{r}1\\4\\-1\\1\end{array}\right],\:\left[\begin{array}{r}2\\-1\\-2\\2\end{array}\right],\:\left[\begin{array}{r}1\\4\\0\\1\end{array}\right]\nonumber\] These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Here are some vectors in \(\mathbb{R}^4\). \[\left[\begin{array}{r}1\\3\\-1\\1\end{array}\right],\:\left[\begin{array}{r}1\\4\\-1\\1\end{array}\right],\:\left[\begin{array}{r}1\\0\\-1\\1\end{array}\right],\:\left[\begin{array}{r}2\\-1\\-2\\2\end{array}\right],\:\left[\begin{array}{r}1\\4\\0\\1\end{array}\right]\nonumber\] These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors

Here are some vectors in \(\mathbb{R}^4\). \[\left[\begin{array}{r}1\\4\\-2\\1\end{array}\right],\:\left[\begin{array}{r}1\\5\\-3\\1\end{array}\right],\:\left[\begin{array}{r}1\\1\\1\\1\end{array}\right],\:\left[\begin{array}{r}2\\1\\3\\2\end{array}\right],\:\left[\begin{array}{r}1\\5\\-2\\1\end{array}\right]\nonumber\] These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Here are some vectors in \(\mathbb{R}^4\). \[\left[\begin{array}{r}1\\-1\\3\\1\end{array}\right],\:\left[\begin{array}{r}1\\0\\7\\1\end{array}\right],\:\left[\begin{array}{r}1\\0\\8\\1\end{array}\right],\:\left[\begin{array}{r}4\\-9\\-6\\4\end{array}\right],\:\left[\begin{array}{r}1\\0\\8\\1\end{array}\right]\nonumber\] These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Here are some vectors in \(\mathbb{R}^4\). \[\left[\begin{array}{r}1\\-1\\-1\\1\end{array}\right],\:\left[\begin{array}{r}-3\\3\\3\\-3\end{array}\right],\:\left[\begin{array}{r}1\\0\\-1\\1\end{array}\right],\:\left[\begin{array}{r}2\\-9\\-2\\2\end{array}\right],\:\left[\begin{array}{r}1\\0\\0\\1\end{array}\right]\nonumber\] These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Here are some vectors in \(\mathbb{R}^4\). \[\left[\begin{array}{r}1\\b+1\\a\\1\end{array}\right],\:\left[\begin{array}{r}3\\3b+3\\3a\\3\end{array}\right],\:\left[\begin{array}{r}1\\b+2\\2a+1\\1\end{array}\right],\:\left[\begin{array}{r}2\\2b-5\\-5a-7\\2\end{array}\right],\:\left[\begin{array}{r}1\\b+2\\2a+2\\1\end{array}\right]\nonumber\] These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Let \(H=span\left\{\left[\begin{array}{r}2\\1\\1\\1\end{array}\right],\:\left[\begin{array}{r}-1\\0\\-1\\-1\end{array}\right],\:\left[\begin{array}{r}5\\2\\3\\3\end{array}\right],\:\left[\begin{array}{r}-1\\1\\-2\\-2\end{array}\right]\right\}\). Find the dimension of \(H\) and determine a basis.

Let \(H\) denote \(span\left\{\left[\begin{array}{r}0\\1\\1\\-1\end{array}\right],\:\left[\begin{array}{r}-1\\-1\\-2\\2\end{array}\right],\:\left[\begin{array}{r}2\\3\\5\\-5\end{array}\right],\:\left[\begin{array}{r}0\\1\\2\\-2\end{array}\right]\right\}\). Find the dimension of \(H\) and determine a basis.

Let \(H\) denote \(span\left\{\left[\begin{array}{r}-2\\1\\1\\-3\end{array}\right],\:\left[\begin{array}{r}-9\\4\\3\\-9\end{array}\right],\:\left[\begin{array}{r}-33\\15\\12\\-36\end{array}\right],\:\left[\begin{array}{r}-22\\10\\8\\-24\end{array}\right]\right\}\). Find the dimension of \(H\) and determine a basis.

Let \(H\) denote \(span\left\{\left[\begin{array}{r}-1\\1\\-1\\-2\end{array}\right],\:\left[\begin{array}{r}-4\\3\\-2\\-4\end{array}\right],\:\left[\begin{array}{r}-3\\2\\-1\\-2\end{array}\right],\:\left[\begin{array}{r}-1\\1\\-2\\-4\end{array}\right],\:\left[\begin{array}{r}-7\\5\\-3\\-6\end{array}\right]\right\}\). Find the dimension of \(H\) and determine a basis.

Let \(H\) denote \(span\left\{\left[\begin{array}{r}2\\3\\2\\1\end{array}\right],\:\left[\begin{array}{r}8\\15\\6\\3\end{array}\right],\:\left[\begin{array}{r}3\\6\\2\\1\end{array}\right],\:\left[\begin{array}{r}4\\6\\6\\3\end{array}\right],\:\left[\begin{array}{r}8\\15\\6\\3\end{array}\right]\right\}\). Find the dimension of \(H\) and determine a basis.

Let \(H\) denote \(span\left\{\left[\begin{array}{r}0\\2\\0\\-1\end{array}\right],\:\left[\begin{array}{r}-1\\6\\0\\-2\end{array}\right],\:\left[\begin{array}{r}-2\\16\\0\\-6\end{array}\right],\:\left[\begin{array}{r}-3\\22\\0\\-8\end{array}\right]\right\}\). Find the dimension of \(H\) and determine a basis.

Let \(H\) denote \(span\left\{\left[\begin{array}{r}5\\1\\1\\4\end{array}\right],\:\left[\begin{array}{r}14\\3\\2\\8\end{array}\right],\:\left[\begin{array}{r}38\\8\\6\\24\end{array}\right],\:\left[\begin{array}{r}47\\10\\7\\28\end{array}\right],\:\left[\begin{array}{r}10\\2\\3\\12\end{array}\right]\right\}\). Find the dimension of \(H\) and determine a basis.

Let \(H\) denote \(span\left\{\left[\begin{array}{r}6\\1\\1\\5\end{array}\right],\:\left[\begin{array}{r}17\\3\\2\\10\end{array}\right],\:\left[\begin{array}{r}52\\9\\7\\35\end{array}\right],\:\left[\begin{array}{r}18\\3\\4\\20\end{array}\right]\right\}\). Find the dimension of \(H\) and determine a basis.

Let \(M=\left\{\vec{u}=\left[\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right]\in\mathbb{R}^4:\sin(u_1)=1\right\}\). Is \(M\) a subspace? Explain.

**Answer**-
No. Let \(\vec{u}=\left[\begin{array}{c}\frac{\pi}{2} \\ 0\\0\\0\end{array}\right]\). Then \(2\vec{u}\cancel{\in}M\) although \(\vec{u}\in M\).

Let \(M=\left\{\vec{u}=\left[\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right]\in\mathbb{R}^4:||u_1||\leq 4\right\}\). Is \(M\) a subspace? Explain.

**Answer**-
No. \(\left[\begin{array}{c}1\\0\\0\\0\end{array}\right]\in M\) but \(10\left[\begin{array}{c}1\\0\\0\\0\end{array}\right]\cancel{\in }M\).

Let \(M=\left\{\vec{u}=\left[\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right]\in\mathbb{R}^4:u_1\geq 0\text{ for each }i=1,2,3,4 \right\}\). Is \(M\) a subspace? Explain.

**Answer**-
This is not a subspace. \(\left[\begin{array}{c}1\\1\\1\\1\end{array}\right]\) is in it. However, \((-1)\left[\begin{array}{c}1\\1\\1\\1\end{array}\right]\) is not.

Let \(\vec{w}\), \(\vec{w}_1\) be given vectors in \(\mathbb{R}^4\) and define \[M=\left\{\vec{u}=\left[\begin{array}{c}u_1\\u_2\\u_3\\u_4\end{array}\right]\in\mathbb{R}^4 :\vec{w}\bullet\vec{u}=0\text{ and }\vec{w}_1\bullet\vec{u}=0\right\}.\nonumber\] Is \(M\) a subspace? Explain.

**Answer**-
This is a subspace because it is closed with respect to vector addition and scalar multiplication.

Let \(\vec{w}\in\mathbb{R}^4\) and let \(M=\left\{\vec{u}=\left[\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right]\in\mathbb{R}^4:\vec{w}\bullet\vec{u}=0\right\}\). Is \(M\) a subspace? Explain.

**Answer**-
Yes, this is a subspace because it is closed with respect to vector addition and scalar multiplication.

Let \(M=\left\{\vec{u}=\left[\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right]\in\mathbb{R}^4:u_3\geq u_1\right\}\). Is \(M\) a subspace? Explain.

**Answer**-
This is not a subspace. \(\left[\begin{array}{c}0\\0\\1\\0\end{array}\right]\) is in it. However \((-1)\left[\begin{array}{c}0\\0\\1\\0\end{array}\right]=\left[\begin{array}{r}0\\0\\-1\\0\end{array}\right]\) is not.

Let \(M=\left\{\vec{u}=\left[\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right]\in\mathbb{R}^4:u_3=u_1=0\right\}\). Is \(M\) a subspace? Explain.

**Answer**-
This is a subspace. It is closed with respect to vector addition and scalar multiplication.

Consider the set of vectors \(S\) given by \[S=\left\{\left[\begin{array}{c}4u+v-5w \\ 12u+6v-6w \\ 4u+4v+4w\end{array}\right] :u,v,w\in\mathbb{R}\right\}.\nonumber\] Is \(S\) a subspace of \(\mathbb{R}^3\)? If so, explain why, give a basis for the subspace and find its dimension.

Consider the set of vectors \(S\) given by \[S=\left\{\left[\begin{array}{c}2u+6v+7w \\ -3u-9v-12w \\ 2u+6v+6w \\ u+3v+3w \end{array}\right] :u,v,w\in\mathbb{R}\right\}.\nonumber\] Is \(S\) a subspace of \(\mathbb{R}^4\)? If so, explain why, give a basis for the subspace and find its dimension.

Consider the set of vectors \(S\) given by \[S=\left\{\left[\begin{array}{c}2u+v \\ 6v-3u+3w \\ 3v-6u+3w \end{array}\right] :u,v,w\in\mathbb{R}\right\}.\nonumber\] Is this set of vectors a subspace of \(\mathbb{R}^3\)? If so, explain why, give a basis for the subspace and find its dimension.

Consider the vectors of the form \[\left\{\left[\begin{array}{c}2u+v+7w \\ u-2v+w \\ -6v-6w \end{array}\right] :u,v,w\in\mathbb{R}\right\}.\nonumber\] Is this set of vectors a subspace of \(\mathbb{R}^3\)? If so, explain why, give a basis for the subspace and find its dimension.

Consider the vectors of the form \[\left\{\left[\begin{array}{c}3u+v+11w \\ 18u+6v+66w \\ 28u+8v+100w \end{array}\right] :u,v,w\in\mathbb{R}\right\}.\nonumber\] Is this set of vectors a subspace of \(\mathbb{R}^3\)? If so, explain why, give a basis for the subspace and find its dimension.

Consider the vectors of the form \[\left\{\left[\begin{array}{c}3u+v \\ 2w-4u \\ 2w-2v-8u \end{array}\right] :u,v,w\in\mathbb{R}\right\}.\nonumber\] Is this set of vectors a subspace of \(\mathbb{R}^3\)? If so, explain why, give a basis for the subspace and find its dimension.

Consider the set of vectors \(S\) given by \[\left\{\left[\begin{array}{c}u+v+w \\ 2u+2v+4w \\ u+v+w \\ 0 \end{array}\right] :u,v,w\in\mathbb{R}\right\}.\nonumber\] Is \(S\) is a subspace of \(\mathbb{R}^4\)? If so, explain why, give a basis for the subspace and find its dimension.

Consider the set of vectors \(S\) given by \[\left\{\left[\begin{array}{c}v \\ -3u-3w \\ 8u-4v+4w \end{array}\right] :u,v,w\in\mathbb{R}\right\}.\nonumber\] Is \(S\) is a subspace of \(\mathbb{R}^4\)? If so, explain why, give a basis for the subspace and find its dimension.

If you have \(5\) vectors in \(\mathbb{R}^5\) and the vectors are linearly independent, can it always be concluded they span \(\mathbb{R}^5\)? Explain.

**Answer**-
Yes. If not, there would exist a vector not in the span. But then you could add in this vector and obtain a linearly independent set of vectors with more vectors than a basis.

If you have \(6\) vectors in \(\mathbb{R}^5\), is it possible they are linearly independent? Explain.

**Answer**-
They can't be.

Suppose \(A\) is an \(m\times n\) matrix and \(\{\vec{w}_1,\cdots ,\vec{w}_k\}\) is a linearly independent set of vectors in \(A(\mathbb{R}^n ) ⊆ \mathbb{R}^m\). Now suppose \(A\vec{z}_i = \vec{w}_i\). Show \(\{\vec{z}_1 ,\cdots ,\vec{z}_k\}\) is also independent.

**Answer**-
Say \(\sum\limits_{i=1}^k c_i\vec{z}_i=\vec{0}\). Then apply \(A\) to it as follows. \[\sum\limits_{i=1}^k c_aA\vec{z}_i=\sum\limits_{i=1}^kc_i\vec{w}_i=\vec{0}\nonumber\] and so, by linear independence of the \(\vec{w}_i\), it follows that each \(c_i=0\).

Suppose \(V,\: W\) are subspaces of \(\mathbb{R}^n\). Let \(V ∩W\) be all vectors which are in both \(V\) and \(W\). Show that \(V ∩W\) is a subspace also.

**Answer**-
If \(\vec{x},\vec{y} ∈ V ∩W\), then for scalars \(α,β\), the linear combination \(α\vec{x} + β\vec{y}\) must be in both \(V\) and \(W\) since they are both subspaces.

Suppose \(V\) and \(W\) both have dimension equal to \(7\) and they are subspaces of \(\mathbb{R}^{10}\). What are the possibilities for the dimension of \(V ∩W\)? Hint: Remember that a linear independent set can be extended to form a basis.

Suppose \(V\) has dimension \(p\) and \(W\) has dimension \(q\) and they are each contained in a subspace, \(U\) which has dimension equal to \(n\) where \(n > \text{max}(p,q)\). What are the possibilities for the dimension of \(V ∩W\)? Hint: Remember that a linearly independent set can be extended to form a basis.

**Answer**-
Let \(\{x_1,\cdots ,x_k\}\) be a basis for \(V∩W\). Then there is a basis for \(V\) and \(W\) which are respectively \[\{x_1, \cdots ,x_k, y_{k+1},\cdots ,y_p\},\:\{x_1,\cdots ,x_k, z_{k+1},\cdots z_q\}\nonumber\] It follows that you must have \(k+p-k+q-k\leq n\) and so you must have \[p+q-n\leq k\nonumber\]

Suppose \(A\) is an \(m\times n\) matrix and \(B\) is an \(n\times p\) matrix. Show that \[\text{dim}(\text{ker}(AB))\leq\text{dim}(\text{ker}(A))+\text{dim}(\text{ker}(B)).\nonumber\] Consider the subspace, \(B(\mathbb{R}^p )∩\text{ker}(A)\) and suppose a basis for this subspace is \(\{\vec{w}_1,\cdots ,\vec{w}_k\}\). Now suppose \(\{\vec{u}_1,\cdots ,\vec{u}_r\}\) is a basis for \(\text{ker}(B)\). Let \(\{\vec{z}_1,\cdots ,\vec{z}_k\}\) be such that \(B\vec{z}_1 =\vec{w}_i\) and argue that \[\text{ker}(AB)⊆ span\{\vec{u}_1,\cdots ,\vec{u}_r,\vec{z}_1,\cdots ,\vec{z}_k\}.\nonumber\]

**Answer**-
Here is how you do this. Suppose \(AB\vec{x} =\vec{0}\). Then \(B\vec{x} ∈ \text{ker}(A) ∩ B(\mathbb{R}^p)\) and so \(B\vec{x} =\sum\limits_{i=1}^k B\vec{z}_i\) showing that \[\vec{x}-\sum\limits_{i=1}^k\vec{z}_i\in\text{ker}(B)\nonumber\] Consider \(B(\mathbb{R}^p )∩\text{ker}(A)\) and let a basis be \(\{\vec{w}_1,\cdots ,\vec{w}_k\}\). Then each \(\vec{w}_i\) is of the form \(B\vec{z}_i =\vec{w}_i\). Therefore, \(\{\vec{z}_1,\cdots ,\vec{z}_k\}\) is linearly independent and \(AB\vec{z}_i = 0\). Now let \(\{\vec{u}_1,\cdots ,\vec{u}_r\}\) be a basis for \(\text{ker}(B)\). If \(AB\vec{x} =\vec{0}\), then \(B\vec{x} ∈ \text{ker}(A)∩B(\mathbb{R}^p)\) and so \(B\vec{x} =\sum\limits_{i=1}^k c_iB\vec{z}_1\) which implies \[\vec{x}-\sum\limits_{i=1}^k c_i\vec{z}_i\in\text{ker}(B)\nonumber\] and so it is of the form \[\vec{x}-\sum\limits_{i=1}^kc_i\vec{z}_i=\sum\limits_{j=1}^r d_j\vec{u}_j\nonumber\] It follows that if \(AB\vec{x} =\vec{0}\) so that \(\vec{x} ∈ \text{ker}(AB)\), then \[\vec{x}\in span (\vec{z}_1,\cdots ,\vec{z}_k,\vec{u}_1, \cdots ,\vec{u}_r ).\nonumber\] Therefore, \[\begin{aligned}\text{dim}(\text{ker}(AB))&\leq k+r=\text{dim}(B(\mathbb{R}^p)∩\text{ker}(A))+\text{dim}(\text{ker}(B)) \\ &\leq\text{dim}(\text{ker}(A))+\text{dim}(\text{ker}(B))\end{aligned}\]

Show that if \(A\) is an \(m\times n\) matrix, then \(\text{ker}(A)\) is a subspace of \(\mathbb{R}^n\).

**Answer**-
If \(\vec{x}\), \(\vec{y}\in\text{ker}(A)\) then \[A(a\vec{x}+b\vec{y})=aA\vec{x}+bA\vec{y}=a\vec{0}+b\vec{0}=\vec{0}\nonumber\] and so \(\text{ker}(A)\) is closed under linear combinations. Hence it is a subspace.

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrrrrr}1&3&0&-2&0&3 \\ 3&9&1&-7&0&8 \\ 1&3&1&-3&1&-1 \\ 1&3&-1&-1&-2&10\end{array}\right]\nonumber\]

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrrrrr}1&3&0&-2&7&3 \\ 3&9&1&-7&23&8 \\ 1&3&1&-3&9&2 \\ 1&3&-1&-1&5&4\end{array}\right]\nonumber\]

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrrrrr}1&0&3&0&7&0 \\ 3&1&10&0&23&0 \\ 1&1&4&1&7&0 \\ 1&-1&2&-2&9&1\end{array}\right]\nonumber\]

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrr}1&0&3 \\ 3&1&10 \\ 1&1&4 \\ 1&-1&2\end{array}\right]\nonumber\]

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrrrr}0&0&-1&0&1 \\ 1&2&3&-2&-18 \\ 1&2&2&-1&-11 \\ -1&-2&-2&1&11\end{array}\right]\nonumber\]

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrrr}1&0&3&0 \\ 3&1&10&0 \\ -1&1&-2&1 \\ 1&-1&2&-2\end{array}\right]\nonumber\]

Find \(\text{ker}(A)\) for the following matrices.

- \(A=\left[\begin{array}{rr}2&3 \\ 4&6\end{array}\right]\)
- \(A=\left[\begin{array}{rrr}1&0&-1 \\ -1&1&3 \\ 3&2&1\end{array}\right]\)
- \(A=\left[\begin{array}{rrr}2&4&0 \\ 3&6&-2 \\ 1&2&-2\end{array}\right]\)
- \(A=\left[\begin{array}{rrrr}2&-1&3&5 \\ 2&0&1&2 \\ 6&4&-5&-6 \\ 0&2&-4&-6\end{array}\right]\)

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \[\left[\begin{array}{c}\frac{1}{6}\sqrt{2}\sqrt{3} \\ \frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3}\end{array}\right],\: \left[\begin{array}{c}\frac{1}{2}\sqrt{2} \\ 0 \\ \frac{1}{2}\sqrt{2}\end{array}\right],\: \left[\begin{array}{c}-\frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3}\end{array}\right]\nonumber\] If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \[\left[\begin{array}{r}1\\2\\-1\end{array}\right],\:\left[\begin{array}{r}1\\0\\1\end{array}\right],\:\left[\begin{array}{r}-1\\1\\1\end{array}\right]\nonumber\] If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \[\left[\begin{array}{r}1\\-1\\1\end{array}\right],\:\left[\begin{array}{r}2\\1\\-1\end{array}\right],\:\left[\begin{array}{r}0\\1\\1\end{array}\right]\nonumber\] If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \[\left[\begin{array}{r}1\\-1\\1\end{array}\right],\:\left[\begin{array}{r}2\\1\\-1\end{array}\right],\:\left[\begin{array}{r}1\\2\\1\end{array}\right]\nonumber\] If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \[\left[\begin{array}{r}1\\0\\0\\0\end{array}\right],\:\left[\begin{array}{r}0\\1\\-1\\0\end{array}\right],\:\left[\begin{array}{r}0\\0\\0\\1\end{array}\right]\nonumber\] If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

Here are some matrices. Label according to whether they are symmetric, skew symmetric, or orthogonal.

- \(\left[\begin{array}{ccc}1&0&0 \\ 0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \\ 0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{array}\right]\)
- \(\left[\begin{array}{ccc}1&2&-3 \\ 2&1&4 \\ -3&4&7\end{array}\right]\)
- \(\left[\begin{array}{ccc}0&-2&-3 \\ 2&0&-4 \\ 3&4&0\end{array}\right]\)

**Answer**-
- Orthogonal
- Symmetric
- Skew Symmetric

For \(U\) an orthogonal matrix, explain why \(||U\vec{x}|| =||\vec{x}||\) for any vector \(\vec{x}\). Next explain why if \(U\) is an \(n\times n\) matrix with the property that \(||U\vec{x}|| =||\vec{x}||\) for all vectors, \(\vec{x}\), then \(U\) must be orthogonal. Thus the orthogonal matrices are exactly those which preserve length.

**Answer**-
\(||U\vec{x}||^2=U\vec{x}\bullet U\vec{x}=U^TU\vec{x}\bullet\vec{x}=I\vec{x}\bullet\vec{x}=||\vec{x}||^2\). Next suppose distance is preserved by \(U\). Then \[\begin{aligned} (U(\vec{x}+\vec{y}))\bullet (U(\vec{x}+\vec{y}))&=||Ux||^2+||Uy||^2+2(Ux\bullet Uy) \\ &=||\vec{x}||^2+||\vec{y}||^2+2(U^TU\vec{x}\bullet\vec{y})\end{aligned}\] But since \(U\) preserves distances, it is also the case that \[(U(\vec{x}+\vec{y})\bullet U(\vec{x}+\vec{y}))=||\vec{x}||^2+||\vec{y}||^2+2(\vec{x}\bullet\vec{y})\nonumber\] Hence \[\vec{x}\bullet\vec{y}=U^TU\vec{x}\bullet\vec{y}\nonumber\] and so \[((U^TU-I)\vec{x})\bullet\vec{y}=0\nonumber\] Since \(y\) is arbitrary, it follows that \(U^TU-I=0\). Thus \(U\) is orthogonal.

Suppose \(U\) is an orthogonal \(n\times n\) matrix. Explain why \(rank(U) = n\).

**Answer**-
You could observe that \(\text{det}(UU^T)=(\text{det}(U))^2-1\) so \(\text{det}(U)\neq 0\).

Fill in the missing entries to make the matrix orthogonal. \[\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\underline{\;}&\underline{\;} \\ \underline{\;}&\frac{\sqrt{6}}{3}&\underline{\;}\end{array}\right].\nonumber\]

**Answer**-
\[\begin{aligned} &\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&a \\ 0&\frac{\sqrt{6}}{3}&b\end{array}\right]\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&a \\ 0&\frac{\sqrt{6}}{3}&b\end{array}\right]^T \\ =&\left[\begin{array}{ccc} 1&\frac{1}{3}\sqrt{3}a-\frac{1}{3} &\frac{1}{3}\sqrt{3}b-\frac{1}{3} \\ \frac{1}{3}\sqrt{3}a-\frac{1}{3}&a^2+\frac{2}{3}&ab-\frac{1}{3} \\ \frac{1}{3}\sqrt{3}b-\frac{1}{3}&ab-\frac{1}{3}&b^2+\frac{2}{3}\end{array}\right]\end{aligned}\] This requires, \(a=1/\sqrt{3},b=1/\sqrt{3}\). \[\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&1/\sqrt{3} \\ 0&\frac{\sqrt{6}}{3}&1/\sqrt{3}\end{array}\right]\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&1/\sqrt{3} \\ 0&\frac{\sqrt{6}}{3}&1/\sqrt{3}\end{array}\right]^T =\left[\begin{array}{ccc}1&0&0 \\ 0&1&0 \\ 0&0&1\end{array}\right]\nonumber\]

Fill in the missing entries to make the matrix orthogonal. \[\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\underline{\;}&\underline{\;} \\ \underline{\;}&0&\underline{\;}\end{array}\right]\nonumber\]

**Answer**-
\[\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2}\\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&a \\ -\frac{1}{3}&0&b\end{array}\right]\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&a \\ -\frac{1}{3}&0&b\end{array}\right]^T=\left[\begin{array}{ccc}1&\frac{1}{6}\sqrt{2}a-\frac{1}{18}&\frac{1}{6}\sqrt{2}b-\frac{2}{9} \\ \frac{1}{6}\sqrt{2}a-\frac{1}{18}&a^2+\frac{17}{18} &ab-\frac{2}{9} \\ \frac{1}{6}\sqrt{2}b-\frac{2}{9}&ab-\frac{2}{9}&b^2+\frac{1}{9}\end{array}\right]\nonumber\] This requires \(a=\frac{1}{3\sqrt{2}},\:b=\frac{4}{3\sqrt{2}}\). \[\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&\frac{1}{3\sqrt{2}} \\ -\frac{1}{3}&0&\frac{4}{3\sqrt{2}}\end{array}\right]\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&\frac{1}{3\sqrt{2}} \\ -\frac{1}{3}&0&\frac{4}{3\sqrt{2}}\end{array}\right]^T=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\nonumber\]

Fill in the missing entries to make the matrix orthogonal. \[\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\underline{\;} \\ \frac{2}{3}&0&\underline{\;} \\ \underline{\;}&\underline{\;}&\frac{4}{15}\sqrt{5}\end{array}\right]\nonumber\]

**Answer**-
Try \[\begin{aligned}&\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&c \\ \frac{2}{3}&0&d \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&c \\ \frac{2}{3}&0&d \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]^T \\ =&\left[\begin{array}{ccc}c^2+\frac{41}{45} &cd+\frac{2}{9}&\frac{4}{15}\sqrt{5}c-\frac{8}{45} \\ cd+\frac{2}{9}&d^2+\frac{4}{9} &\frac{4}{15}\sqrt{5}d+\frac{4}{9} \\ \frac{4}{15}\sqrt{5}c-\frac{8}{45}&\frac{4}{15}\sqrt{5}d+\frac{4}{9}&1\end{array}\right]\end{aligned}\] This requires that \(c=\frac{2}{3\sqrt{5}},d=\frac{-5}{3\sqrt{5}}\). \[\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\frac{2}{3\sqrt{5}} \\ \frac{2}{3}&0&\frac{-5}{3\sqrt{5}} \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\frac{2}{3\sqrt{5}} \\ \frac{2}{3}&0&\frac{-5}{3\sqrt{5}} \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]^T=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\nonumber\]

Find an orthonormal basis for the span of each of the following sets of vectors.

- \(\left[\begin{array}{r}3\\-4\\0\end{array}\right],\:\left[\begin{array}{r}7\\-1\\0\end{array}\right],\:\left[\begin{array}{r}1\\7\\1\end{array}\right]\)
- \(\left[\begin{array}{r}3\\0\\-4\end{array}\right],\:\left[\begin{array}{r}11\\0\\2\end{array}\right],\:\left[\begin{array}{r}1\\1\\7\end{array}\right]\)
- \(\left[\begin{array}{r}3\\0\\-4\end{array}\right],\:\left[\begin{array}{r}5\\0\\10\end{array}\right],\:\left[\begin{array}{r}-7\\1\\1\end{array}\right]\)

**Answer**-
- \(\left[\begin{array}{c}\frac{3}{5} \\ -\frac{4}{5} \\ 0\end{array}\right],\:\left[\begin{array}{c}\frac{4}{5}\\ \frac{3}{5} \\ 0\end{array}\right],\:\left[\begin{array}{c}0\\0\\1\end{array}\right]\)
- \(\left[\begin{array}{c}\frac{3}{5}\\ 0\\ -\frac{4}{5}\end{array}\right],\:\left[\begin{array}{c}\frac{4}{5} \\ 0\\ \frac{3}{5}\end{array}\right],\:\left[\begin{array}{c}0\\1\\0\end{array}\right]\)
- \(\left[\begin{array}{c}\frac{3}{5}\\0\\-\frac{4}{5}\end{array}\right],\:\left[\begin{array}{c}\frac{4}{5}\\0\\ \frac{3}{5}\end{array}\right],\:\left[\begin{array}{c}0\\1\\0\end{array}\right]\)

Using the Gram Schmidt process find an orthonormal basis for the following span: \[span\left\{\left[\begin{array}{r}1\\2\\1\end{array}\right],\:\left[\begin{array}{r}2\\-1\\3\end{array}\right],\:\left[\begin{array}{r}1\\0\\0\end{array}\right]\right\}\nonumber\]

**Answer**-
A solution is \[\left[\begin{array}{c}\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{6} \\ \frac{1}{6}\sqrt{6}\end{array}\right],\:\left[\begin{array}{c}\frac{3}{10}\sqrt{2} \\ -\frac{2}{5}\sqrt{2} \\ \frac{1}{2}\sqrt{2}\end{array}\right],\:\left[\begin{array}{c}\frac{7}{15}\sqrt{3} \\ -\frac{1}{15}\sqrt{3} \\ -\frac{1}{3}\sqrt{3}\end{array}\right]\nonumber\]

Using the Gram Schmidt process find an orthonormal basis for the following span: \[span\left\{\left[\begin{array}{r}1\\2\\1\\0\end{array}\right],\:\left[\begin{array}{r}2\\-1\\3\\1\end{array}\right],\:\left[\begin{array}{r}1\\0\\0\\1\end{array}\right]\right\}\nonumber\]

**Answer**-
Then a solution is \[\left[\begin{array}{c}\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{6} \\ \frac{1}{6}\sqrt{6} \\ 0\end{array}\right],\:\left[\begin{array}{c}\frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{2}{9}\sqrt{2}\sqrt{3} \\ \frac{5}{18}\sqrt{2}\sqrt{3} \\ \frac{1}{9}\sqrt{2}\sqrt{3}\end{array}\right],\:\left[\begin{array}{c}\frac{5}{111}\sqrt{3}\sqrt{37} \\ \frac{1}{133}\sqrt{3}\sqrt{37} \\ -\frac{17}{333}\sqrt{3}\sqrt{37} \\ \frac{22}{333}\sqrt{3}\sqrt{37}\end{array}\right]\nonumber\]

The set \(V=\left\{\left[\begin{array}{c}x\\y\\z\end{array}\right] :2x+3y-z=0\right\}\) is a subspace of \(\mathbb{R}^3\). Find an orthonormal basis for this subspace.

**Answer**-
The subspace is of the form \[\left[\begin{array}{c}x\\y\\2x+3y\end{array}\right]\nonumber\] and a basis is \(\left[\begin{array}{c}1\\0\\2\end{array}\right],\:\left[\begin{array}{c}0\\1\\3\end{array}\right]\). Therefore, an orthonormal basis is \[\left[\begin{array}{c}\frac{1}{5}\sqrt{5} \\ 0\\ \frac{2}{5}\sqrt{5}\end{array}\right],\:\left[\begin{array}{c}-\frac{3}{35}\sqrt{5}\sqrt{14} \\ \frac{1}{14}\sqrt{5}\sqrt{14} \\ \frac{3}{70}\sqrt{5}\sqrt{14}\end{array}\right]\nonumber\]

Consider the following scalar equation of a plane. \[2x-3y+z=0\nonumber\] Find the orthogonal complement of the vector \(\vec{v}=\left[\begin{array}{c}3\\4\\1\end{array}\right]\). Also find the point on the plane which is closest to \((3,4,1)\).

Consider the following scalar equation of a plane. \[x+3y+z=0\nonumber\] Find the orthogonal complement of the vector \(\vec{v}=\left[\begin{array}{c}1\\2\\1\end{array}\right]\). Also find the point on the plane which is closest to \((3,4,1)\).

Let \(\vec{v}\) be a vector and let \(\vec{n}\) be a normal vector for a plane through the origin. Find the equation of the line through the point determined by \(\vec{v}\) which has direction vector \(\vec{n}\). Show that it intersects the plane at the point determined by \(\vec{v}−proj_{\vec{n}}\vec{v}\). Hint: The line:\(\vec{v}+t\vec{n}\). It is in the plane if \(\vec{n}•(\vec{v}+t\vec{n}) = 0\). Determine \(t\). Then substitute in to the equation of the line.

As shown in the above problem, one can find the closest point to~v in a plane through the origin by finding the intersection of the line through \(\vec{v}\) having direction vector equal to the normal vector to the plane with the plane. If the plane does not pass through the origin, this will still work to find the point on the plane closest to the point determined by \(\vec{v}\). Here is a relation which defines a plane \[2x+y+z=11\nonumber\] and here is a point: \((1, 1, 2)\). Find the point on the plane which is closest to this point. Then determine the distance from the point to the plane by taking the distance between these two points. Hint: Line: \((x, y,z) = (1, 1, 2) +t(2, 1, 1)\). Now require that it intersect the plane.

In general, you have a point \((x_0, y_0,z_0)\) and a scalar equation for a plane \(ax+by+cz = d\) where \(a^2 +b^2 +c^2 > 0\). Determine a formula for the closest point on the plane to the given point. Then use this point to get a formula for the distance from the given point to the plane. Hint: Find the line perpendicular to the plane which goes through the given point: \((x, y,z) = (x_0, y_0,z_0) + t(a,b, c)\). Now require that this point satisfy the equation for the plane to determine \(t\).

Find the least squares solution to the following system. \[\begin{aligned}x+2y&=1 \\ 2x+3y&=2 \\ 3x+5y&=4\end{aligned}\]

**Answer**-
\[\begin{aligned}\left[\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right]^T\left[\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right]&=\left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right] \\ &=\left[\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right]^T\left[\begin{array}{c}1\\2\\4\end{array}\right]=\left[\begin{array}{c}17\\28\end{array}\right]\end{aligned}\] \[\begin{aligned}\left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]&=\left[\begin{array}{c}17\\28\end{array}\right] \\ \left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]&=\left[\begin{array}{c}17\\28\end{array}\right],\end{aligned}\] Solution is: \(\left[\begin{array}{c}\frac{2}{3}\\ \frac{1}{3}\end{array}\right]\)

You are doing experiments and have obtained the ordered pairs, \[(0, 1),(1, 2),(2, 3.5),(3, 4)\nonumber\] Find \(m\) and \(b\) such that \(\vec{y} = m\vec{x}+b\) approximates these four points as well as possible.

Suppose you have several ordered triples, \((x_i , y_i ,z_i)\). Describe how to find a polynomial such as \[z = a+bx+cy+dxy+ex^2 + fy^2\nonumber\] giving the best fit to the given ordered triples.

The wind blows from the South at \(20\) kilometers per hour and an airplane which flies at \(600\) kilometers per hour in still air is heading East. Find the velocity of the airplane and its location after two hours.

The wind blows from the West at \(30\) kilometers per hour and an airplane which flies at \(400\) kilometers per hour in still air is heading North East. Find the velocity of the airplane and its position after two hours.

The wind blows from the North at \(10\) kilometers per hour. An airplane which flies at \(300\) kilometers per hour in still air is supposed to go to the point whose coordinates are at \(\left( 100, 100 \right).\) In what direction should the airplane fly?

Three forces act on an object. Two are \(\left [ \begin{array}{r} 3 \\ -1 \\ -1 \end{array} \right ]\) and \(\left [ \begin{array}{r} 1 \\ -3 \\ 4 \end{array} \right ]\) Newtons. Find the third force if the object is not to move.

Three forces act on an object. Two are \(\left [ \begin{array}{r} 6 \\ -3 \\ 3 \end{array} \right ]\) and \(\left [ \begin{array}{r} 2 \\ 1 \\ 3 \end{array} \right ]\) Newtons. Find the third force if the total force on the object is to be \(\left [ \begin{array}{r} 7 \\ 1 \\ 3 \end{array} \right ] .\)

A river flows West at the rate of \(b\) miles per hour. A boat can move at the rate of \(8\) miles per hour. Find the smallest value of \(b\) such that it is not possible for the boat to proceed directly across the river.

The wind blows from West to East at a speed of \(50\) miles per hour and an airplane which travels at \(400\) miles per hour in still air is heading North West. What is the velocity of the airplane relative to the ground? What is the component of this velocity in the direction North?

**Answer**-
The velocity is the sum of two vectors. \(50\vec{i}+\frac{ 300}{\sqrt{2}} \left( \vec{i}+\vec{j}\right) =\left( 50+\frac{300}{\sqrt{2}} \right) \vec{i}+ \frac{300}{\sqrt{2}}\vec{j}.\) The component in the direction of North is then \(\frac{300}{\sqrt{2}}= 150\sqrt{2}\) and the velocity relative to the ground is \[\left( 50+\frac{300}{\sqrt{2}}\right) \vec{i}+\frac{300}{\sqrt{2}}\vec{j}\nonumber \]

The wind blows from West to East at a speed of \(60\) miles per hour and an airplane can travel travels at \(100\) miles per hour in still air. How many degrees West of North should the airplane head in order to travel exactly North?

The wind blows from West to East at a speed of \(50\) miles per hour and an airplane which travels at \(400\) miles per hour in still air heading somewhat West of North so that, with the wind, it is flying due North. It uses \(30.0\) gallons of gas every hour. If it has to travel \(600.0\) miles due North, how much gas will it use in flying to its destination?

An airplane is flying due north at \(150.0\) miles per hour but it is not actually going due North because there is a wind which is pushing the airplane due east at \(40.0\) miles per hour. After one hour, the plane starts flying \(30^{\circ }\) East of North. Assuming the plane starts at \(\left( 0,0\right) ,\) where is it after \(2\) hours? Let North be the direction of the positive \(y\) axis and let East be the direction of the positive \(x\) axis.

**Answer**-
Velocity of plane for the first hour: \(\left [ \begin{array}{cc} 0 & 150 \end{array} \right ] + \left [ \begin{array}{cc} 40 & 0 \end{array} \right ] =\left [ \begin{array}{cc} 40 & 150 \end{array} \right ] .\) After one hour it is at \(\left( 40,150\right) .\) Next the velocity of the plane is \(150\left [ \begin{array}{cc} \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array} \right ] +\left [ \begin{array}{cc} 40 & 0 \end{array} \right ]\) in miles per hour. After two hours it is then at \(\left( 40,150\right) + 150\left [ \begin{array}{cc} \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array} \right ] +\left [ \begin{array}{cc} 40 & 0 \end{array} \right ] = \left [ \begin{array}{cc} 155 & 75\sqrt{3}+150 \end{array} \right ] = \left [ \begin{array}{cc} 155.0 & 279.\, 9 \end{array} \right ]\)

City A is located at the origin \(\left( 0,0 \right)\) while city B is located at \(\left(300,500 \right)\) where distances are in miles. An airplane flies at \(250\) miles per hour in still air. This airplane wants to fly from city A to city B but the wind is blowing in the direction of the positive \(y\) axis at a speed of \(50\) miles per hour. Find a unit vector such that if the plane heads in this direction, it will end up at city B having flown the shortest possible distance. How long will it take to get there?

**Answer**-
Wind: \(\left [ \begin{array}{cc} 0 & 50 \end{array} \right ] .\) Direction it needs to travel: \(\left( 3,5 \right) \frac{1}{\sqrt{34}}.\) Then you need \(250 \left [ \begin{array}{cc} a & b \end{array} \right ] + \left [ \begin{array}{cc} 0 & 50 \end{array} \right ]\) to have this direction where \(\left [ \begin{array}{cc} a & b \end{array} \right ]\) is an appropriate unit vector. Thus you need \[\begin{aligned} a^{2}+b^{2} &=1 \\ \frac{250b+50}{250a} &=\frac{5}{3}\end{aligned}\] Thus \(a=\frac{3}{5},b=\frac{4}{5}.\) The velocity of the plane relative to the ground is \(\left [ \begin{array}{cc} 150 & 250 \end{array} \right ] .\) The speed of the plane relative to the ground is given by \[\sqrt{\left( 150\right) ^{2}+\left( 250\right) ^{2}}= 291.55 \text{ miles per hour }\nonumber\] It has to go a distance of \(\sqrt{\left( 300\right) ^{2}+\left( 500\right) ^{2}}= 583.\, 10\) miles. Therefore, it takes \[\frac{ 583.\, 1}{ 291.\, 55}=2 \text{ hours}\nonumber \]

A certain river is one half mile wide with a current flowing at \(2\) miles per hour from East to West. A man swims directly toward the opposite shore from the South bank of the river at a speed of \(3\) miles per hour. How far down the river does he find himself when he has swam across? How far does he end up traveling?

**Answer**-
Water:\(\left [ \begin{array}{rr} -2 & 0 \end{array} \right ]\) Swimmer:\(\left [ \begin{array}{rr} 0 & 3 \end{array} \right ]\) Speed relative to earth: \(\left [ \begin{array}{rr} -2 & 3 \end{array} \right ] .\) It takes him \(1/6\) of an hour to get across. Therefore, he ends up traveling \(\frac{1}{6}\sqrt{4+9}= \frac{1}{6}\sqrt{13}\) miles. He ends up \(1/3\) mile down stream.

A certain river is one half mile wide with a current flowing at 2 miles per hour from East to West. A man can swim at \(3\) miles per hour in still water. In what direction should he swim in order to travel directly across the river? What would the answer to this problem be if the river flowed at 3 miles per hour and the man could swim only at the rate of 2 miles per hour?

**Answer**-
Man: \(3\left [ \begin{array}{rr} a & b \end{array} \right ]\) Water: \(\left [ \begin{array}{rr} -2 & 0 \end{array} \right ]\) Then you need \(3a=2\) and so \(a=2/3\) and hence \(b=\sqrt{5}/3\). The vector is then \(\left [ \begin{array}{cc} \frac{2}{3} & \frac{\sqrt{5}}{3} \end{array} \right ] .\)

In the second case, he could not do it. You would need to have a unit vector \(\left [ \begin{array}{rr} a & b \end{array} \right ]\) such that \(2a=3\) which is not possible.

Three forces are applied to a point which does not move. Two of the forces are \(2 \vec{i}+2 \vec{j} -6 \vec{k}\) Newtons and \(8 \vec{i}+ 8 \vec{j}+ 3 \vec{k}\) Newtons. Find the third force.

The total force acting on an object is to be \(4 \vec{i}+ 2 \vec{j} -3 \vec{k}\) Newtons. A force of \(-3 \vec{i} -1 \vec{j}+ 8 \vec{k}\) Newtons is being applied. What other force should be applied to achieve the desired total force?

A bird flies from its nest \(8\) km in the direction \(\frac{5}{6}\pi\) north of east where it stops to rest on a tree. It then flies \(1\) km in the direction due southeast and lands atop a telephone pole. Place an \(xy\) coordinate system so that the origin is the bird’s nest, and the positive \(x\) axis points east and the positive \(y\) axis points north. Find the displacement vector from the nest to the telephone pole.

If \(\vec{F}\) is a force and \(\vec{D}\) is a vector, show \(\mathrm{proj}_{\vec{D}}\left( \vec{F}\right) =\left( \| \vec{F} \| \cos \theta \right) \vec{u}\) where \(\vec{u}\) is the unit vector in the direction of \(\vec{D}\), where \(\vec{u}=\vec{D}/ \| \vec{D} \|\) and \(\theta\) is the included angle between the two vectors, \(\vec{F}\) and \(\vec{D}\). \( \| \vec{F} \| \cos \theta\) is sometimes called the component of the force, \(\vec{F}\) in the direction, \(\vec{D}\).

**Answer**-
\(\mathrm{proj}_{\vec{D}}\left( \vec{F}\right) = \frac{\vec{F}\bullet \vec{D}}{ \| \vec{D} \| }\frac{\vec{D}}{ \| \vec{D} \| }=\left( \| \vec{F} \| \cos \theta \right) \frac{\vec{D}}{ \| \vec{D} \| }=\left( \| \vec{F} \| \cos \theta \right) \vec{u}\)

A boy drags a sled for \(100\) feet along the ground by pulling on a rope which is \(20\) degrees from the horizontal with a force of \(40\) pounds. How much work does this force do?

**Answer**-
\(40\cos \left( \frac{20}{180}\pi \right)100=3758.8\)

A girl drags a sled for \(200\) feet along the ground by pulling on a rope which is \(30\) degrees from the horizontal with a force of \(20\) pounds. How much work does this force do?

**Answer**-
\(20\cos \left( \frac{\pi }{6}\right)200= 3464.1\)

A large dog drags a sled for \(300\) feet along the ground by pulling on a rope which is \(45\) degrees from the horizontal with a force of \(20\) pounds. How much work does this force do?

**Answer**-
\(20\left( \cos \frac{\pi }{4}\right)300=4242.6\)

How much work does it take to slide a crate \(20\) meters along a loading dock by pulling on it with a \(200\) Newton force at an angle of \(30^{\circ }\) from the horizontal? Express your answer in Newton meters.

**Answer**-
\(200\left( \cos \left( \frac{\pi }{6}\right) \right) 20= 3464.1\)

An object moves \(10\) meters in the direction of \(\vec{j}\). There are two forces acting on this object, \(\vec{F}_{1}=\vec{i}+\vec{j}+ 2\vec{k}\), and \(\vec{F}_{2}=-5\vec{i}+2\vec{j}-6\vec{k}\). Find the total work done on the object by the two forces. **Hint:** You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?

**Answer**-
\(\left [ \begin{array}{r} -4 \\ 3 \\ -4 \end{array} \right ] \bullet \left [ \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right ] \times 10= 30\) You can consider the resultant of the two forces because of the properties of the dot product.

An object moves \(10\) meters in the direction of \(\vec{j}+\vec{i}\). There are two forces acting on this object, \(\vec{F}_{1}=\vec{i}+2\vec{j} +2\vec{k}\), and \(\vec{F}_{2}=5\vec{i}+2\vec{j}-6\vec{k}\). Find the total work done on the object by the two forces. **Hint:** You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?

**Answer**-
\[\begin{aligned} \vec{F}_{1}\bullet \left [ \begin{array}{r} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \right ] 10+\vec{F}_{2}\bullet \left [ \begin{array}{r} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \right ] 10 &=\left( \vec{F}_{1}+\vec{F}_{2}\right) \bullet \left [ \begin{array}{r} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \right ] 10 \\ &= \left [ \begin{array}{r} 6 \\ 4 \\ -4 \end{array} \right ] \bullet \left [ \begin{array}{r} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \right ] 10 \\ &= 50\sqrt{2}\end{aligned}\]

An object moves \(20\) meters in the direction of \(\vec{k}+\vec{j}\). There are two forces acting on this object, \(\vec{F}_{1}=\vec{i}+\vec{j}+ 2\vec{k}\), and \(\vec{F}_{2}=\vec{i}+2\vec{j}-6\vec{k}\). Find the total work done on the object by the two forces. **Hint:** You can take the work done by the resultant of the two forces or you can add the work done by each force.

**Answer**-
\(\left [ \begin{array}{r} 2 \\ 3 \\ -4 \end{array} \right ] \bullet \left [ \begin{array}{r} 0 \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right ] 20= -10\sqrt{2}\)