4.3E: More on the Cross Product Exercises

Exercises for 1

solutions

2

If \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) are the coordinate vectors, verify that \(\mathbf{i} \times \mathbf{j} = \mathbf{k}\), \(\mathbf{j} \times \mathbf{k} = \mathbf{i}\), and \(\mathbf{k} \times \mathbf{i} = \mathbf{j}\).

Show that \(\mathbf{u} \times (\mathbf{v} \times \mathbf{w})\) need not equal \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}\) by calculating both when

\[\mathbf{u} = \left[ \begin{array}{r} 1\\ 1\\ 1 \end{array} \right], \mathbf{v} = \left[ \begin{array}{r} 1\\ 1\\ 0 \end{array} \right], \mbox{ and } \mathbf{w} = \left[ \begin{array}{r} 0\\ 0\\ 1 \end{array} \right] \nonumber \]

Find two unit vectors orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\) if:

1. \(\mathbf{u} = \left[ \begin{array}{r} 1\\ 2\\ 2 \end{array} \right]\), \(\mathbf{v} = \left[ \begin{array}{r} 2\\ -1\\ 2 \end{array} \right]\)
2. \(\mathbf{u} = \left[ \begin{array}{r} 1\\ 2\\ -1 \end{array} \right]\), \(\mathbf{v} = \left[ \begin{array}{r} 3\\ 1\\ 2 \end{array} \right]\)

2. \(\pm \frac{\sqrt{3}}{3} \left[ \begin{array}{r} 1\\ -1\\ -1 \end{array} \right]\).

Find the area of the triangle with the following vertices.

1. \(A(3, -1, 2)\), \(B(1, 1, 0)\), and \(C(1, 2, -1)\)
2. \(A(3, 0, 1)\), \(B(5, 1, 0)\), and \(C(7, 2, -1)\)
3. \(A(1, 1, -1)\), \(B(2, 0, 1)\), and \(C(1, -1, 3)\)
4. \(A(3, -1, 1)\), \(B(4, 1, 0)\), and \(C(2, -3, 0)\)

2. \(0\)
3. \(\sqrt{5}\)

Find the volume of the parallelepiped determined by \(\mathbf{w}\), \(\mathbf{u}\), and \(\mathbf{v}\) when:

1. \(\mathbf{w} = \left[ \begin{array}{r} 2\\ 1\\ 1 \end{array} \right]\), \(\mathbf{v} = \left[ \begin{array}{r} 1\\ 0\\ 2 \end{array} \right]\), and \(\mathbf{u} = \left[ \begin{array}{r} 2\\ 1\\ -1 \end{array} \right]\)
2. \(\mathbf{w} = \left[ \begin{array}{r} 1\\ 0\\ 3 \end{array} \right]\), \(\mathbf{v} = \left[ \begin{array}{r} 2\\ 1\\ -3 \end{array} \right]\), and \(\mathbf{u} = \left[ \begin{array}{r} 1\\ 1\\ 1 \end{array} \right]\)

2. \(7\)

Let \(P_{0}\) be a point with vector \(\mathbf{p}_{0}\), and let \(ax + by + cz = d\) be the equation of a plane with normal \(\mathbf{n} = \left[ \begin{array}{r} a\\ b\\ c \end{array} \right]\).

1. Show that the point on the plane closest to \(P_{0}\) has vector \(\mathbf{p}\) given by

   \[\mathbf{p} = \mathbf{p}_{0} + \frac{d - (\mathbf{p}_{0}\bullet \mathbf{n})}{\| \mathbf{n} \|^2}\mathbf{n}. \nonumber \]

2. Show that the shortest distance from \(P_{0}\) to the plane is \(\frac{|d - (\mathbf{p}_{0}\bullet \mathbf{n})|}{\| \mathbf{n} \|}\).
3. Show that \(\mathbf{p}_{0} + 2\frac{d - (\mathbf{p}_{0}\bullet \mathbf{n})}{\| \mathbf{n} \|^2}\mathbf{n}\) is the vector of \(P_{0}^\prime\).

2. The distance is \(\|\mathbf{p} - \mathbf{p}_{0}\|\); use part (a.).

Simplify \((a\mathbf{u} + b\mathbf{v}) \times (c\mathbf{u} + d\mathbf{v})\).

Show that the shortest distance from a point \(P\) to the line through \(P_{0}\) with direction vector \(\mathbf{d}\) is \(\frac{\| \longvect{P_{0}P} \times \mathbf{d} \|}{\| \mathbf{d} \|}\).

Let \(\mathbf{u}\) and \(\mathbf{v}\) be nonzero, nonorthogonal vectors. If \(\theta\) is the angle between them, show that \(\tan\theta = \frac{\| \mathbf{u} \times \mathbf{v} \|}{\mathbf{u}\bullet \mathbf{v}}.\)

Show that points \(A\), \(B\), and \(C\) are all on one line if and only if \(\longvect{AB} \times \longvect{AC} = 0\)

\(\| \longvect{AB} \times \longvect{AC} \|\) is the area of the parallelogram determined by \(A\), \(B\), and \(C\).

Show that points \(A\), \(B\), \(C\), and \(D\) are all on one plane if and only if \(\longvect{AB}\bullet (\longvect{AB} \times \longvect{AC}) = 0\)

Use Theorem [thm:012765] to confirm that, if \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) are mutually perpendicular, the (rectangular) parallelepiped they determine has volume \(\|\mathbf{u}\|\|\mathbf{v}\|\|\mathbf{w}\|\).

Because \(\mathbf{u}\) and \(\mathbf{v} \times \mathbf{w}\) are parallel, the angle \(\theta\) between them is \(0\) or \(\pi\). Hence \(\cos(\theta) = \pm 1\), so the volume is \(|\mathbf{u}\bullet (\mathbf{v} \times \mathbf{w})| = \|\mathbf{u}\|\|\mathbf{v} \times \mathbf{w}\| \cos(\theta) = \|\mathbf{u}\|\|(\mathbf{v} \times \mathbf{w})\|\). But the angle between \(\mathbf{v}\) and \(\mathbf{w}\) is \(\frac{\pi}{2}\) so \(\|\mathbf{v} \times \mathbf{w}\| = \|\mathbf{v} \| \|\mathbf{w}\| \cos(\frac{\pi}{2}) = \|\mathbf{v}\|\|\mathbf{w}\|\). The result follows.

Show that the volume of the parallelepiped determined by \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{u} \times \mathbf{v}\) is \(\|\mathbf{u} \times \mathbf{v}\|^{2}\).

[ex:ch4_3_ex14] Complete the proof of Theorem [thm:012715].

[ex:ch4_3_ex15] Prove the following properties in Theorem [thm:012690].

Property 6 Property 7 Property 8

2. If \(\mathbf{u} = \left[ \begin{array}{r} u_{1}\\ u_{2}\\ u_{3} \end{array} \right]\), \(\mathbf{v} = \left[ \begin{array}{r} v_{1}\\ v_{2}\\ v_{3} \end{array} \right]\) and \(\mathbf{w} = \left[ \begin{array}{r} w_{1}\\ w_{2}\\ w_{3} \end{array} \right]\), then \(\mathbf{u} \times (\mathbf{v} + \mathbf{w}) = \det \left[ \begin{array}{rrr} \mathbf{i} & u_{1} & v_{1} + w_{1}\\ \mathbf{j} & u_{2} & v_{2} + w_{2}\\ \mathbf{k} & u_{3} & v_{3} + w_{3} \end{array} \right]\)
   \({} = \det \left[ \begin{array}{rrr} \mathbf{i} & u_{1} & v_{1}\\ \mathbf{j} & u_{2} & v_{2}\\ \mathbf{k} & u_{3} & v_{3} \end{array} \right] + \det \left[ \begin{array}{rrr} \mathbf{i} & u_{1} & w_{1}\\ \mathbf{j} & u_{2} & w_{2}\\ \mathbf{k} & u_{3} & w_{3} \end{array} \right]\)
   \({} = (\mathbf{u} \times \mathbf{v}) + (\mathbf{u} \times \mathbf{w})\) where we used Exercise [ex:ch4_3_ex21].

[ex:ch4_3_ex16]

1. Show that \(\mathbf{w}\bullet (\mathbf{u} \times \mathbf{v}) = \mathbf{u}\bullet (\mathbf{v} \times \mathbf{w}) = \mathbf{v} \times (\mathbf{w} \times \mathbf{u})\) holds for all vectors \(\mathbf{w}\), \(\mathbf{u}\), and \(\mathbf{v}\).
2. Show that \(\mathbf{v} - \mathbf{w}\) and \((\mathbf{u} \times \mathbf{v}) + (\mathbf{v} \times \mathbf{w}) + (\mathbf{w} \times \mathbf{u})\) are orthogonal.

2. \((\mathbf{v} - \mathbf{w})\bullet [(\mathbf{u} \times \mathbf{v}) + (\mathbf{v} \times \mathbf{w}) + (\mathbf{w} \times \mathbf{u})] = (\mathbf{v} - \mathbf{w})\bullet (\mathbf{u} \times \mathbf{v}) + (\mathbf{v} - \mathbf{w})\bullet (\mathbf{v} \times \mathbf{w}) + (\mathbf{v} - \mathbf{w})\bullet (\mathbf{w} \times \mathbf{u}) = -\mathbf{w}\bullet (\mathbf{u} \times \mathbf{v}) + 0 + \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) = 0\).

[ex:ch4_3_ex17] Show \(\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u}\bullet \mathbf{w}) \mathbf{v} - (\mathbf{u} \times \mathbf{v})\mathbf{w}\). [Hint: First do it for \(\mathbf{u} = \mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\); then write \(\mathbf{u} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) and use Theorem [thm:012690].]

Prove the Jacobi identity:

\[\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) + \mathbf{v} \times (\mathbf{w} \times \mathbf{u}) + \mathbf{w} \times (\mathbf{u} \times \mathbf{v}) = \mathbf{0} \nonumber \]

[Hint: The preceding exercise.]

Show that

\[(\mathbf{u} \times \mathbf{v})\bullet (\mathbf{w} \times \mathbf{z}) = \det \left[ \begin{array}{cc} \mathbf{u}\bullet \mathbf{w} & \mathbf{u}\bullet \mathbf{z}\\ \mathbf{v}\bullet \mathbf{w} & \mathbf{v}\bullet \mathbf{z} \end{array} \right] \nonumber \]

[Hint: Exercises [ex:ch4_3_ex16] and [ex:ch4_3_ex17].]

Let \(P\), \(Q\), \(R\), and \(S\) be four points, not all on one plane, as in the diagram. Show that the volume of the pyramid they determine is

\[\frac{1}{6}|\longvect{PQ}\bullet (\longvect{PR} \times \longvect{PS})|. \nonumber \]

[Hint: The volume of a cone with base area \(A\) and height \(h\) as in the diagram below right is \(\frac{1}{3}Ah\).]

[ex:ch4_3_ex21] Consider a triangle with vertices \(A\), \(B\), and \(C\), as in the diagram below. Let \(\alpha\), \(\beta\), and \(\gamma\) denote the angles at \(A\), \(B\), and \(C\), respectively, and let \(a\), \(b\), and \(c\) denote the lengths of the sides opposite \(A\), \(B\), and \(C\), respectively. Write \(\mathbf{u} = \longvect{AB}\), \(\mathbf{v} = \longvect{BC}\), and \(\mathbf{w} = \longvect{CA}\).

1. Deduce that \(\mathbf{u} + \mathbf{v} + \mathbf{w} = \mathbf{0}\).
2. Show that \(\mathbf{u} \times \mathbf{v} = \mathbf{w} \times \mathbf{u} = \mathbf{v} \times \mathbf{w}\). [Hint: Compute \(\mathbf{u} \times (\mathbf{u} + \mathbf{v} + \mathbf{w})\) and \(\mathbf{v} \times (\mathbf{u} + \mathbf{v} + \mathbf{w})\).]
3. Deduce the law of sines:

   \[\frac{\sin\alpha}{a} = \frac{\sin\beta}{b} = \frac{\sin\gamma}{c} \nonumber \]

Show that the (shortest) distance between two planes \(\mathbf{n}\bullet \mathbf{p} = d_{1}\) and \(\mathbf{n}\bullet \mathbf{p} = d_{2}\) with \(\mathbf{n}\) as normal is \(\frac{|d_{2} - d_{1}|}{\| \mathbf{n} \|}\).

Let \(\mathbf{p}_{1}\) and \(\mathbf{p}_{2}\) be vectors of points in the planes, so \(\mathbf{p}_{1}\bullet \mathbf{n} = d_{1}\) and \(\mathbf{p}_{2}\bullet \mathbf{n} = d_{2}\). The distance is the length of the projection of \(\mathbf{p}_{2} - \mathbf{p}_{1}\) along \(\mathbf{n}\); that is \(\frac{|(\mathbf{p}_{2} - \mathbf{p}_{1})\bullet \mathbf{n}|}{\| \mathbf{n} \|} = \frac{|d_{1} - d_{2}|}{\| \mathbf{n} \|}\).

Let \(A\) and \(B\) be points other than the origin, and let \(\mathbf{a}\) and \(\mathbf{b}\) be their vectors. If \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel, show that the plane through \(A\), \(B\), and the origin is given by

\[\{P(x, y, z) \mid \left[ \begin{array}{r} x\\ y\\ z \end{array} \right] = s\mathbf{a} + t\mathbf{b} \mbox{ for some } s \mbox{ and } t \} \nonumber \]

Let \(A\) be a \(2 \times 3\) matrix of rank 2 with rows \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\). Show that

\[P = \{XA \mid X = [x y]; x, y \mbox{ arbitrary}\} \nonumber \]

is the plane through the origin with normal \(\mathbf{r}_{1} \times \mathbf{r}_{2}\).

Given the cube with vertices \(P(x, y, z)\), where each of \(x\), \(y\), and \(z\) is either \(0\) or \(2\), consider the plane perpendicular to the diagonal through \(P(0, 0, 0)\) and \(P(2, 2, 2)\) and bisecting it.

1. Show that the plane meets six of the edges of the cube and bisects them.
2. Show that the six points in (a) are the vertices of a regular hexagon.