4.1E: Vectors and Lines Exercises

Exercises for 1

solutions

2

Compute $\|\mathbf{v}\|$ if $\mathbf{v}$ equals:

$\left[ \begin{array}{r} 2 \\ -1 \\ 2 \end{array} \right]$ $\left[ \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right]$ $\left[ \begin{array}{r} 1 \\ 0 \\ -1 \end{array} \right]$ $\left[ \begin{array}{r} -1 \\ 0 \\ 2 \end{array} \right]$ $2\left[ \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right]$ $-3\left[ \begin{array}{r} 1 \\ 1 \\ 2 \end{array} \right]$

2. $\sqrt{6}$
3. $\sqrt{5}$
4. $3\sqrt{6}$

Find a unit vector in the direction of:

$\left[ \begin{array}{r} 7 \\ -1 \\ 5 \end{array} \right]$ $\left[ \begin{array}{r} -2 \\ -1 \\ 2 \end{array} \right]$

2. $\frac{1}{3}\left[ \begin{array}{r} -2 \\ -1 \\ 2 \end{array} \right]$

1. Find a unit vector in the direction from
   $\left[ \begin{array}{r} 3 \\ -1 \\ 4 \end{array} \right]$ to $\left[ \begin{array}{r} 1\\ 3 \\ 5 \end{array} \right]$.

2. If $\mathbf{u} \neq \mathbf{0}$, for which values of $a$ is $a\mathbf{u}$ a unit vector?

Find the distance between the following pairs of points.

$\left[ \begin{array}{r} 3 \\ -1 \\ 0 \end{array} \right]$ and $\left[ \begin{array}{r} 2\\ -1 \\ 1 \end{array} \right]$ $\left[ \begin{array}{r} 2 \\ -1 \\ 2 \end{array} \right]$ and $\left[ \begin{array}{r} 2\\ 0 \\ 1 \end{array} \right]$ $\left[ \begin{array}{r} -3 \\ 5 \\ 2 \end{array} \right]$ and $\left[ \begin{array}{r} 1\\ 3 \\ 3 \end{array} \right]$ $\left[ \begin{array}{r} 4 \\ 0 \\ -2 \end{array} \right]$ and $\left[ \begin{array}{r} 3\\ 2 \\ 0 \end{array} \right]$

2. $\sqrt{2}$
3. $3$

Use vectors to show that the line joining the midpoints of two sides of a triangle is parallel to the third side and half as long.

Let $A$, $B$, and $C$ denote the three vertices of a triangle.

1. If $E$ is the midpoint of side $BC$, show that

   $$\longvect{AE} = \frac{1}{2}(\longvect{AB} + \longvect{AC}) \nonumber$$

2. If $F$ is the midpoint of side $AC$, show that

   $$\longvect{FE} = \frac{1}{2}\longvect{AB} \nonumber$$

2. $\longvect{FE} = \longvect{FC} + \longvect{CE} = \frac{1}{2}\longvect{AC} + \frac{1}{2}\longvect{CB} = \frac{1}{2}(\longvect{AC} + \longvect{CB}) = \frac{1}{2}\longvect{AB}$

Determine whether $\mathbf{u}$ and $\mathbf{v}$ are parallel in each of the following cases.

1. $\mathbf{u} = \left[ \begin{array}{r} -3\\ -6\\ 3 \end{array} \right]$; $\mathbf{v} = \left[ \begin{array}{r} 5\\ 10 \\ -5 \end{array} \right]$
2. $\mathbf{u} = \left[ \begin{array}{r} 3\\ -6\\ 3 \end{array} \right]$; $\mathbf{v} = \left[ \begin{array}{r} -1\\ 2 \\ -1 \end{array} \right]$
3. $\mathbf{u} = \left[ \begin{array}{r} 1\\ 0\\ 1 \end{array} \right]$; $\mathbf{v} = \left[ \begin{array}{r} -1\\ 0 \\ 1 \end{array} \right]$
4. $\mathbf{u} = \left[ \begin{array}{r} 2\\ 0\\ -1 \end{array} \right]$; $\mathbf{v} = \left[ \begin{array}{r} -8\\ 0 \\ 4 \end{array} \right]$

2. Yes
3. Yes

Let $\mathbf{p}$ and $\mathbf{q}$ be the vectors of points $P$ and $Q$, respectively, and let $R$ be the point whose vector is $\mathbf{p} + \mathbf{q}$. Express the following in terms of $\mathbf{p}$ and $\mathbf{q}$.

$\longvect{QP}$ $\longvect{QR}$ $\longvect{RP}$ $\longvect{RO}$ where $O$ is the origin

2. $\mathbf{p}$
3. $-(\mathbf{p} + \mathbf{q})$.

In each case, find $\longvect{PQ}$ and $\| \longvect{PQ} \|$.

1. $P(1, -1, 3)$, $Q(3, 1, 0)$
2. $P(2, 0, 1)$, $Q(1, -1, 6)$
3. $P(1, 0, 1)$, $Q(1, 0, -3)$
4. $P(1, -1, 2)$, $Q(1, -1, 2)$
5. $P(1, 0, -3)$, $Q(-1, 0, 3)$
6. $P(3, -1, 6)$, $Q(1, 1, 4)$

2. $\left[ \begin{array}{r} -1\\ -1\\ 5 \end{array} \right]$, $\sqrt{27}$
3. $\left[ \begin{array}{r} 0\\ 0\\ 0 \end{array} \right]$, $0$
4. $\left[ \begin{array}{r} -2\\ 2\\ 2 \end{array} \right]$, $\sqrt{12}$

In each case, find a point $Q$ such that $\longvect{PQ}$ has (i) the same direction as $\mathbf{v}$; (ii) the opposite direction to $\mathbf{v}$.

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

2. (i) $Q(5, -1, 2)$ (ii) $Q(1, 1, -4)$.

Let $\mathbf{u} = \left[ \begin{array}{r} 3\\ -1\\ 0 \end{array} \right]$, $\mathbf{v} = \left[ \begin{array}{r} 4\\ 0\\ 1 \end{array} \right]$, and $\mathbf{w} = \left[ \begin{array}{r} -1\\ 1\\ 5 \end{array} \right]$. In each case, find $\mathbf{x}$ such that:

1. $3(2\mathbf{u} + \mathbf{x}) + \mathbf{w} = 2\mathbf{x} - \mathbf{v}$
2. $2(3\mathbf{v} - \mathbf{x}) = 5\mathbf{w} + \mathbf{u} - 3\mathbf{x}$

2. $\mathbf{x} = \mathbf{u} - 6\mathbf{v} + 5\mathbf{w} = \left[ \begin{array}{r} -26\\ 4\\ 19 \end{array} \right]$

Let $\mathbf{u} = \left[ \begin{array}{r} 1\\ 1\\ 2 \end{array} \right]$, $\mathbf{v} = \left[ \begin{array}{r} 0\\ 1\\ 2 \end{array} \right]$, and $\mathbf{w} = \left[ \begin{array}{r} 1\\ 0\\ -1 \end{array} \right]$. In each case, find numbers $a$, $b$, and $c$ such that $\mathbf{x} = a\mathbf{u} + b\mathbf{v} + c\mathbf{w}$.

$\mathbf{x} = \left[ \begin{array}{r} 2\\ -1\\ 6 \end{array} \right]$ $\mathbf{x} = \left[ \begin{array}{r} 1\\ 3\\ 0 \end{array} \right]$

2. $\left[ \begin{array}{r} a\\ b\\ c \end{array} \right] = \left[ \begin{array}{r} -5\\ 8\\ 6 \end{array} \right]$

Let $\mathbf{u} = \left[ \begin{array}{r} 3\\ -1\\ 0 \end{array} \right]$, $\mathbf{v} = \left[ \begin{array}{r} 4\\ 0\\ 1 \end{array} \right]$, and $\mathbf{z} = \left[ \begin{array}{r} 1\\ 1\\ 1 \end{array} \right]$. In each case, show that there are no numbers $a$, $b$, and $c$ such that:

1. $a\mathbf{u} + b\mathbf{v} + c\mathbf{z} = \left[ \begin{array}{r} 1\\ 2\\ 1 \end{array} \right]$
2. $a\mathbf{u} + b\mathbf{v} + c\mathbf{z} = \left[ \begin{array}{r} 5\\ 6\\ -1 \end{array} \right]$

2. If it holds then $\left[ \begin{array}{c} 3a + 4b + c\\ -a + c\\ b + c \end{array} \right] = \left[ \begin{array}{c} x_{1}\\ x_{2}\\ x_{3} \end{array} \right]$.
   $\left[ \begin{array}{rrrr} 3 & 4 & 1 & x_{1}\\ -1 & 0 & 1 & x_{2}\\ 0 & 1 & 1 & x_{3} \end{array} \right] \to \left[ \begin{array}{rrrc} 0 & 4 & 4 & x_{1} + 3x_{2}\\ -1 & 0 & 1 & x_{2}\\ 0 & 1 & 1 & x_{3} \end{array} \right]$

   If there is to be a solution then $x_{1} + 3x_{2} = 4x_{3}$ must hold. This is not satisfied.

Given $P_{1}(2, 1, -2)$ and $P_{2}(1, -2, 0)$. Find the coordinates of the point $P$:

1. $\frac{1}{5}$ the way from $P_{1}$ to $P_{2}$
2. $\frac{1}{4}$ the way from $P_{2}$ to $P_{1}$

2. $\frac{1}{4}\left[ \begin{array}{c} 5\\ -5\\ -2 \end{array} \right]$

Find the two points trisecting the segment between $P(2, 3, 5)$ and $Q(8, -6, 2)$.

Let $P_{1}(x_{1}, y_{1}, z_{1})$ and $P_{2}(x_{2}, y_{2}, z_{2})$ be two points with vectors $\mathbf{p}_{1}$ and $\mathbf{p}_{2}$, respectively. If $r$ and $s$ are positive integers, show that the point $P$ lying $\frac{r}{r + s}$ the way from $P_{1}$ to $P_{2}$ has vector

$$\mathbf{p} = \left( \frac{s}{r + s} \right)\mathbf{p}_{1} + \left( \frac{r}{r + s} \right)\mathbf{p}_{2} \nonumber$$

In each case, find the point $Q$:

1. $\longvect{PQ} = \left[ \begin{array}{r} 2\\ 0\\ -3 \end{array} \right]$ and $P = P(2,-3,1)$
2. $\longvect{PQ} = \left[ \begin{array}{r} -1\\ 4\\ 7 \end{array} \right]$ and $P = P(1,3,-4)$

2. $Q(0, 7, 3)$.

Let $\mathbf{u} = \left[ \begin{array}{r} 2\\ 0\\ -4 \end{array} \right]$ and $\mathbf{v} = \left[ \begin{array}{r} 2\\ 1\\ -2 \end{array} \right]$. In each case find $\mathbf{x}$:

1. $2\mathbf{u} - \| \mathbf{v} \| \mathbf{v} = \frac{3}{2}(\mathbf{u} - 2\mathbf{x})$
2. $3\mathbf{u} + 7\mathbf{v} = \|\mathbf{u}\|^{2}(2\mathbf{x} + \mathbf{v})$

2. $\mathbf{x} = \frac{1}{40}\left[ \begin{array}{r} -20\\ -13\\ 14 \end{array} \right]$

Find all vectors $\mathbf{u}$ that are parallel to $\mathbf{v} = \left[ \begin{array}{r} 3\\ -2\\ 1 \end{array} \right]$ and satisfy $\|\mathbf{u}\| = 3\|\mathbf{v}\|$.

Let $P$, $Q$, and $R$ be the vertices of a parallelogram with adjacent sides $PQ$ and $PR$. In each case, find the other vertex $S$.

1. $P(3, -1, -1)$, $Q(1, -2, 0)$, $R(1, -1, 2)$
2. $P(2, 0, -1)$, $Q(-2, 4, 1)$, $R(3, -1, 0)$

2. $S(-1, 3, 2)$.

In each case either prove the statement or give an example showing that it is false.

1. The zero vector $\mathbf{0}$ is the only vector of length 0.
2. If $\|\mathbf{v} - \mathbf{w}\| = 0$, then $\mathbf{v} = \mathbf{w}$.
3. If $\mathbf{v} = -\mathbf{v}$, then $\mathbf{v} = \mathbf{0}$.
4. If $\|\mathbf{v}\| = \|\mathbf{w}\|$, then $\mathbf{v} = \mathbf{w}$.
5. If $\|\mathbf{v}\| = \|\mathbf{w}\|$, then $\mathbf{v} = \pm\mathbf{w}$.
6. If $\mathbf{v} = t\mathbf{w}$ for some scalar $t$, then $\mathbf{v}$ and $\mathbf{w}$ have the same direction.
7. If $\mathbf{v}$, $\mathbf{w}$, and $\mathbf{v} + \mathbf{w}$ are nonzero, and $\mathbf{v}$ and $\mathbf{v} + \mathbf{w}$ parallel, then $\mathbf{v}$ and $\mathbf{w}$ are parallel.
8. $\|-5\mathbf{v}\| = -5\|\mathbf{v}\|$, for all $\mathbf{v}$.
9. If $\|\mathbf{v}\| = \| 2\mathbf{v}\|$, then $\mathbf{v} = \mathbf{0}$.
10. $\|\mathbf{v} + \mathbf{w}\| = \|\mathbf{v}\| + \|\mathbf{w}\|$, for all $\mathbf{v}$ and $\mathbf{w}$.

2. T. $\|\mathbf{v} - \mathbf{w}\| = 0$ implies that $\mathbf{v} - \mathbf{w} = \mathbf{0}$.
3. F. $\|\mathbf{v}\| = \| - \mathbf{v}\|$ for all $\mathbf{v}$ but $\mathbf{v} = -\mathbf{v}$ only holds if $\mathbf{v} = \mathbf{0}$.
4. F. If $t < 0$ they have the opposite direction.
5. F. $\| -5\mathbf{v}\| = 5\|\mathbf{v}\|$ for all $\mathbf{v}$, so it fails if $\mathbf{v} \neq \mathbf{0}$.
6. F. Take $\mathbf{w} = -\mathbf{v}$ where $\mathbf{v} \neq \mathbf{0}$.

Find the vector and parametric equations of the following lines.

1. The line parallel to $\left[ \begin{array}{r} 2\\ -1\\ 0 \end{array} \right]$ and passing through $P(1, -1, 3)$.
2. The line passing through $P(3, -1, 4)$ and $Q(1, 0, -1)$.
3. The line passing through $P(3, -1, 4)$ and $Q(3, -1, 5)$.
4. The line parallel to $\left[ \begin{array}{r} 1\\ 1\\ 1 \end{array} \right]$ and passing through $P(1, 1, 1)$.
5. The line passing through $P(1, 0, -3)$ and parallel to the line with parametric equations $x = -1 + 2t$, $y = 2 - t$, and $z = 3 + 3t$.
6. The line passing through $P(2, -1, 1)$ and parallel to the line with parametric equations $x = 2 - t$, $y = 1$, and $z = t$.
7. The lines through $P(1, 0, 1)$ that meet the line with vector equation $\mathbf{p} = \left[ \begin{array}{r} 1\\ 2\\ 0 \end{array} \right] + t \left[ \begin{array}{r} 2\\ -1\\ 2 \end{array} \right]$ at points at distance 3 from $P_{0}(1, 2, 0)$.

2. $\left[ \begin{array}{r} 3\\ -1\\ 4 \end{array} \right] + t \left[ \begin{array}{r} 2\\ -1\\ 5 \end{array} \right]$; $x = 3 + 2t$, $y = -1 -t$, $z = 4 + 5t$
3. $\left[ \begin{array}{r} 1\\ 1\\ 1 \end{array} \right] + t \left[ \begin{array}{r} 1\\ 1\\ 1 \end{array} \right]$; $x = y = z = 1 + t$
4. $\left[ \begin{array}{r} 2\\ -1\\ 1 \end{array} \right] + t \left[ \begin{array}{r} -1\\ 0\\ 1 \end{array} \right]$; $x = 2 - t$, $y = -1$, $z = 1 + t$

In each case, verify that the points $P$ and $Q$ lie on the line.

1. $\begin{array}[t]{ll} x = 3 - 4t & P(-1,3,0), Q(11,0,3) \\ y = 2 + t & \\ z = 1 - t & \end{array}$
2. $\begin{array}[t]{ll} x = 4 - t & P(2,3,-3), Q(-1,3,-9) \\ y = 3 & \\ z = 1 - 2t & \end{array}$

2. $P$ corresponds to $t = 2$; $Q$ corresponds to $t = 5$.

Find the point of intersection (if any) of the following pairs of lines.

1. $\begin{array}[t]{ll} x = 3 + t & x = 4 + 2s \\ y = 1 - 2t & y = 6 + 3s \\ z = 3 + 3t & z = 1 + s \end{array}$
2. $\begin{array}{ll} x = 1 - t & x = 2s \\ y = 2 + 2t & y = 1 + s \\ z = -1 + 3t & z = 3 \end{array}$
3. $\left[ \begin{array}{c} x\\ y\\ z \end{array} \right] = \left[ \begin{array}{r} 3\\ -1\\ 2 \end{array} \right] + t \left[ \begin{array}{r} 1\\ 1\\ -1 \end{array} \right]$

   $\left[ \begin{array}{c} x\\ y\\ z \end{array} \right] = \left[ \begin{array}{r} 1\\ 1\\ -2 \end{array} \right] + s \left[ \begin{array}{r} 2\\ 0\\ 3 \end{array} \right]$

4. $\left[ \begin{array}{c} x\\ y\\ z \end{array} \right] = \left[ \begin{array}{r} 4\\ -1\\ 5 \end{array} \right] + t \left[ \begin{array}{r} 1\\ 0\\ 1 \end{array} \right]$

   $\left[ \begin{array}{c} x\\ y\\ z \end{array} \right] = \left[ \begin{array}{r} 2\\ -7\\ 12 \end{array} \right] + s \left[ \begin{array}{r} 0\\ -2\\ 3 \end{array} \right]$

2. No intersection
3. $P(2, -1, 3)$; $t = -2$, $s = -3$

Show that if a line passes through the origin, the vectors of points on the line are all scalar multiples of some fixed nonzero vector.

Show that every line parallel to the $z$ axis has parametric equations $x = x_{0}$, $y = y_{0}$, $z = t$ for some fixed numbers $x_{0}$ and $y_{0}$.

Let $\mathbf{d} = \left[ \begin{array}{c} a\\ b\\ c \end{array} \right]$ be a vector where $a$, $b$, and $c$ are all nonzero. Show that the equations of the line through $P_{0}(x_{0}, y_{0}, z_{0})$ with direction vector $\mathbf{d}$ can be written in the form

$$\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z -z_{0}}{c} \nonumber$$

This is called the symmetric form of the equations.

A parallelogram has sides $AB$, $BC$, $CD$, and $DA$. Given $A(1, -1, 2)$, $C(2, 1, 0)$, and the midpoint $M(1, 0, -3)$ of $AB$, find $\longvect{BD}$.

Find all points $C$ on the line through $A(1, -1, 2)$ and $B = (2, 0, 1)$ such that $\| \longvect{AC} \| = 2 \| \longvect{BC} \|$.

$P(3, 1, 0)$ or $P(\frac{5}{3}, \frac{-1}{3}, \frac{4}{3})$

Let $A$, $B$, $C$, $D$, $E$, and $F$ be the vertices of a regular hexagon, taken in order. Show that $\longvect{AB} + \longvect{AC} + \longvect{AD} + \longvect{AE} + \longvect{AF} = 3\longvect{AD}$.

1. Let $P_{1}$, $P_{2}$, $P_{3}$, $P_{4}$, $P_{5}$, and $P_{6}$ be six points equally spaced on a circle with centre $C$. Show that

   $$\longvect{CP}_{1} + \longvect{CP}_{2} + \longvect{CP}_{3} + \longvect{CP}_{4} + \longvect{CP}_{5} + \longvect{CP}_{6} = \mathbf{0} \nonumber$$

2. Show that the conclusion in part (a) holds for any even set of points evenly spaced on the circle.
3. Show that the conclusion in part (a) holds for three points.
4. Do you think it works for any finite set of points evenly spaced around the circle?

2. $\longvect{CP}_{k} = -\longvect{CP}_{n+k}$ if $1 \leq k \leq n$, where there are $2n$ points.

Consider a quadrilateral with vertices $A$, $B$, $C$, and $D$ in order (as shown in the diagram).

If the diagonals $AC$ and $BD$ bisect each other, show that the quadrilateral is a parallelogram. (This is the converse of Example [exa:011062].) [Hint: Let $E$ be the intersection of the diagonals. Show that $\longvect{AB} = \longvect{DC}$ by writing $\longvect{AB} = \longvect{AE} + \longvect{EB}$.]

Consider the parallelogram $ABCD$ (see diagram), and let $E$ be the midpoint of side $AD$.

Show that $BE$ and $AC$ trisect each other; that is, show that the intersection point is one-third of the way from $E$ to $B$ and from $A$ to $C$. [Hint: If $F$ is one-third of the way from $A$ to $C$, show that $2\longvect{EF} = \longvect{FB}$ and argue as in Example [exa:011062].]

$\longvect{DA} = 2\longvect{EA}$ and $2\longvect{AF} = \longvect{FC}$, so $2\longvect{EF} = 2(\longvect{EF} + \longvect{AF}) = \longvect{DA} + \longvect{FC} = \longvect{CB} + \longvect{FC} = \longvect{FC} + \longvect{CB} = \longvect{FB}$. Hence $\longvect{EF} = \frac{1}{2}\longvect{FB}$. So $F$ is the trisection point of both $AC$ and $EB$.

The line from a vertex of a triangle to the midpoint of the opposite side is called a median of the triangle. If the vertices of a triangle have vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$, show that the point on each median that is $\frac{1}{3}$ the way from the midpoint to the vertex has vector $\frac{1}{3}(\mathbf{u} + \mathbf{v} + \mathbf{w})$. Conclude that the point $C$ with vector $\frac{1}{3}(\mathbf{u} + \mathbf{v} + \mathbf{w})$ lies on all three medians. This point $C$ is called the centroid of the triangle.

Given four noncoplanar points in space, the figure with these points as vertices is called a tetrahedron. The line from a vertex through the centroid (see previous exercise) of the triangle formed by the remaining vertices is called a median of the tetrahedron. If $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$, and $\mathbf{x}$ are the vectors of the four vertices, show that the point on a median one-fourth the way from the centroid to the vertex has vector $\frac{1}{4}(\mathbf{u} + \mathbf{v} + \mathbf{w} + \mathbf{x})$. Conclude that the four medians are concurrent.