4.3E: Exercises for Section 4.2

Exercise \(\PageIndex{1}\)

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

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

Exercise \(\PageIndex{2}\)

Use the formula given in Proposition 4.2.1 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.

Exercise \(\PageIndex{3}\)

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

Exercise \(\PageIndex{4}\)

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

Exercise \(\PageIndex{5}\)

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

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

Exercise \(\PageIndex{6}\)

Use the result of Problem \(\PageIndex{5}\) 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 \(B^TA^T-(AB)^T=0\).

Exercise \(\PageIndex{7}\)

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.

Exercise \(\PageIndex{8}\)

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.

Exercise \(\PageIndex{9}\)

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

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

Exercise \(\PageIndex{10}\)

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

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

Exercise \(\PageIndex{11}\)

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

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

Exercise \(\PageIndex{12}\)

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 \(P\).

Exercise \(\PageIndex{13}\)

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 \(P\).

Exercise \(\PageIndex{14}\)

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.

Exercise \(\PageIndex{15}\)

The Cauchy Schwarz inequality in \(\mathbb{R}^n\) is:

For vectors \(\vec{u}\), \(\vec{v}\), \((\vec{u}-\text{proj}_{\vec{v}}\vec{u})\bullet (\vec{u}-\text{proj}_{\vec{v}}\vec{u})\geq 0\).

1. Prove the Cauchy Schwarz Inequality. HINT: Begin by simplifying using the axioms of the dot product and then putting 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 \(\vec{u} = \text{proj}_{\vec{v}}\vec{u}\).
2. What is the geometric meaning of \(\vec{w}= \text{proj}_{\vec{v}}\vec{w}\)?

Answer

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

Exercise \(\PageIndex{16}\)

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

Exercise \(\PageIndex{17}\)

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 \(\vec{u}\).

Exercise \(\PageIndex{18}\)

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.

Exercise \(\PageIndex{19}\)

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 \(9\).

Exercise \(\PageIndex{20}\)

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 \[\frac{1}{2}\sqrt{1+(18)^2+49}=\frac{1}{2}\sqrt{374}\nonumber\]

Exercise \(\PageIndex{21}\)

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.

Exercise \(\PageIndex{22}\)

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 \(8\sqrt{3}\).

Exercise \(\PageIndex{23}\)

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 \(\sqrt{36+121+4}=\sqrt{161}\).

Exercise \(\PageIndex{24}\)

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.

Exercise \(\PageIndex{25}\)

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. \[\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\1&0&0\\0&1&0\end{array}\right|=\vec{k}\nonumber\]

Exercise \(\PageIndex{26}\)

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 \(\omega_1\vec{i}+\omega_2\vec{j}+\omega_3\vec{k}\).

Exercise \(\PageIndex{27}\)

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

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

Exercise \(\PageIndex{28}\)

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.

Exercise \(\PageIndex{29}\)

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.