VECTOR ALGEBRA

Chapter 10

Vector Algebra

Big idea: some quantities need a direction as well as a size. Vectors give them an algebra, so that pushes, journeys and velocities can be added, scaled and compared with simple arithmetic.

Think of two kinds of information. "The room is $4$ metres long" needs only a number. "Walk $4$ metres" is incomplete: which way? Some quantities are complete with a number and a unit. Others also need a direction.

• A scalar has only a size (magnitude). Examples: mass, time, temperature, distance, speed, area, energy.
• A vector has a size and a direction. Examples: displacement, velocity, acceleration, force, momentum.

A person walks $3$ blocks east and then $4$ blocks north. The distance walked is $3 + 4 = 7$ blocks: a scalar, which only counts how far the feet moved. The displacement is the straight line from the start to the end. It is a vector. By Pythagoras its size is $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$ blocks, and its direction is north of east.

Vectors are used wherever direction matters: in physics (forces, velocities), engineering (structures, circuits), computer graphics (movement, lighting) and navigation.

Every point in space is described by three coordinates $(x, y, z)$, measured along three perpendicular axes that meet at the origin $O = (0, 0, 0)$. The $z$-axis points up. The distance between two points comes from Pythagoras' theorem, used twice:

$$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

Example: for $P = (1, 2, 3)$ and $Q = (4, -2, 5)$, the differences are $3$, $-4$, $2$. The squares add to $29$, so $PQ = \sqrt{29}$.

A vector is a directed line segment: a straight line with a start and an end, drawn with an arrowhead at the end. If it starts at $A$ and ends at $B$, we write $\overrightarrow{AB}$. The start $A$ is the initial point and the end $B$ is the terminal point. A vector is also written with a single letter and an arrow, such as $\vec{a}$.

The magnitude of $\overrightarrow{AB}$ is the length of the segment, written $|\overrightarrow{AB}|$. It equals the distance $AB$. Two things describe a vector completely: its magnitude and its direction. It does not matter where on the page it is drawn.

Here $|\overrightarrow{AB}| = \sqrt{17} \approx 4.1231$.

If you fix the origin $O$, every point $P$ gets its own vector $\overrightarrow{OP}$ from the origin to $P$. This is the position vector of $P$. If $P = (x, y, z)$, then

$$\overrightarrow{OP} = x\,\hat{i} + y\,\hat{j} + z\,\hat{k}.$$

Here $\hat{i}$, $\hat{j}$, $\hat{k}$ are unit vectors (length $1$) along the $x$, $y$ and $z$ axes. The numbers $x$, $y$, $z$ are the components of the vector. The magnitude comes from the distance formula:

$$|\overrightarrow{OP}| = \sqrt{x^2 + y^2 + z^2}.$$

Example 1. For $P = (2, 3, 6)$: $\overrightarrow{OP} = 2\hat{i} + 3\hat{j} + 6\hat{k}$ and $|\overrightarrow{OP}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

Example 2. For $P = (1, -2, 2)$: $\overrightarrow{OP} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $|\overrightarrow{OP}| = \sqrt{1 + 4 + 4} = 3$.

A vector $\vec{r}$ makes three angles with the positive $x$, $y$, $z$ axes. Call them $\alpha$, $\beta$, $\gamma$ (the direction angles). The cosines of these angles are the direction cosines:

$$l = \cos\alpha, \qquad m = \cos\beta, \qquad n = \cos\gamma.$$

For $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ of length $r$, the $x$-component is the adjacent side of the angle with the $x$-axis, so

$$l = \frac{x}{r}, \qquad m = \frac{y}{r}, \qquad n = \frac{z}{r}.$$

Square and add: $l^2 + m^2 + n^2 = \dfrac{x^2 + y^2 + z^2}{r^2} = 1$. This always holds. The direction cosines are also the components of the unit vector $\hat{r} = l\hat{i} + m\hat{j} + n\hat{k}$.

For $\vec{r} = 2\hat{i} + 3\hat{j} + 6\hat{k}$ we have $r = 7$. The direction cosines are $\frac{2}{7}$, $\frac{3}{7}$, $\frac{6}{7}$. Their squares are $\frac{4}{49}, \frac{9}{49}, \frac{36}{49}$, which add to $\frac{49}{49} = 1$. The direction angles are $\alpha \approx 73.4^{\circ}$, $\beta \approx 64.62^{\circ}$, $\gamma \approx 31^{\circ}$.

For $\vec{r} = \hat{i} - 2\hat{j} + 2\hat{k}$ we have $r = 3$. The direction cosines are $\frac{1}{3}$, $-\frac{2}{3}$, $\frac{2}{3}$. The negative sign tells us the vector leans away from the positive $y$-axis, so $\beta \approx 131.81^{\circ}$ is more than $90^{\circ}$. The other angles are $\alpha \approx 70.53^{\circ}$ and $\gamma \approx 48.19^{\circ}$.

Any numbers proportional to the direction cosines are called direction ratios. For $\vec{r}$ above, $(2, 3, 6)$ and $(4, 6, 12)$ are both direction ratios, because the second is the first times $2$. Direction cosines are the special direction ratios whose squares add to $1$. To get them from direction ratios $a, b, c$, divide each by $\sqrt{a^2 + b^2 + c^2}$.

If a vector makes equal angles with all three axes, then $l = m = n$. From $l^2 + m^2 + n^2 = 1$ we get $3l^2 = 1$, so $l = \pm\dfrac{1}{\sqrt{3}}$. The vector $\hat{i} + \hat{j} + \hat{k}$ is one such vector, and its angle with each axis is $\cos^{-1}\dfrac{1}{\sqrt{3}} \approx 54.74^{\circ}$.

Dividing a vector by its own length gives a unit vector in the same direction. We write it $\hat{a} = \dfrac{\vec{a}}{|\vec{a}|}$. For $\vec{a} = 3\hat{i} + 4\hat{j} + 12\hat{k}$ the length is $\sqrt{9 + 16 + 144} = 13$. So

$$\hat{a} = \frac{3}{13}\hat{i} + \frac{4}{13}\hat{j} + \frac{12}{13}\hat{k}.$$

As a check, $\frac{9 + 16 + 144}{169} = 1$, and the decimals are $0.2308, 0.3077, 0.9231$.

Two vectors are collinear if one is a number times the other. For $\vec{a} = 2\hat{i} - 4\hat{j} + 6\hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j} - 3\hat{k}$, we see that $\vec{a} = -2\vec{b}$. So they are collinear. The minus sign means they point in opposite directions.

Take $P = (0, 1, 1)$ and $Q = (1, 3, 3)$. The vector $\overrightarrow{PQ}$ goes $1$ along $x$, $2$ along $y$, $2$ along $z$, so $\overrightarrow{PQ} = \hat{i} + 2\hat{j} + 2\hat{k}$. It is equal to the vector $\hat{i} + 2\hat{j} + 2\hat{k}$ drawn from the origin, even though it starts somewhere else.

1. Say whether each quantity is a scalar or a vector: temperature, displacement, area, force, volume, acceleration, kinetic energy, momentum.
2. Find the magnitude of each position vector: $3\hat{i} + 4\hat{j} + 12\hat{k}$, $-2\hat{i} + 6\hat{j} + 9\hat{k}$, $\hat{i} + 2\hat{j} + 2\hat{k}$, $-5\hat{j} + 12\hat{k}$.
3. Find the direction cosines of $\vec{r} = 6\hat{i} - 2\hat{j} + 3\hat{k}$ and the angle it makes with the $x$-axis.
4. Find the unit vector in the direction of $4\hat{i} - 3\hat{j} + 12\hat{k}$.
5. Which of $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$, $\vec{b} = -2\hat{i} + 4\hat{j} - 6\hat{k}$, $\vec{c} = 3\hat{i} - 6\hat{j} + 9\hat{k}$ are collinear? Which pairs point the same way?
6. Are $\hat{i} + 2\hat{j} + 3\hat{k}$ and $2\hat{i} + 4\hat{j} + 5\hat{k}$ collinear?
7. Find $x$ so that $x\hat{i} - 2\hat{j} + 4\hat{k}$ is collinear with $3\hat{i} - 6\hat{j} + 12\hat{k}$.
8. Write the vector from $A = (1, 2, 3)$ to $B = (4, 6, 3)$ and find its magnitude.
9. Find the magnitude of $3\vec{a} - 2\vec{b}$ for $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$.
10. Explain why $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} + \hat{j} + 0\hat{k}$ are the same vector, and give its direction cosines.

Answers

1. Scalars: temperature, area, volume, kinetic energy. Vectors: displacement, force, acceleration, momentum.
2. $13$, $11$, $3$, $13$. For example $\sqrt{9 + 16 + 144} = \sqrt{169} = 13$.
3. $|\vec{r}| = 7$, so the direction cosines are $\frac{6}{7}$, $-\frac{2}{7}$, $\frac{3}{7}$. The angle with the $x$-axis is $\cos^{-1}\frac{6}{7} \approx 31^{\circ}$.
4. The length is $\sqrt{16 + 9 + 144} = 13$, so the unit vector is $\dfrac{4}{13}\hat{i} - \dfrac{3}{13}\hat{j} + \dfrac{12}{13}\hat{k}$.
5. All three are collinear: $\vec{b} = -2\vec{a}$ and $\vec{c} = 3\vec{a}$. The vectors $\vec{a}$ and $\vec{c}$ point the same way (positive multiple). $\vec{b}$ points opposite to both.
6. No. The first two components of the second are twice those of the first, but the third is not ($5 \ne 6$), so it is not a multiple of the first.
7. The second vector is $3$ times $\hat{i} - 2\hat{j} + 4\hat{k}$, so $x = 1$.
8. $\overrightarrow{AB} = 3\hat{i} + 4\hat{j}$ and $|\overrightarrow{AB}| = \sqrt{9 + 16 + 0} = 5$.
9. $3\vec{a} = 3\hat{i} + 6\hat{j} - 3\hat{k}$ and $2\vec{b} = 4\hat{i} - 2\hat{j} + 6\hat{k}$. Their difference is $-\hat{i} + 8\hat{j} - 9\hat{k}$, with magnitude $\sqrt{1 + 64 + 81} = \sqrt{146}$.
10. The component of $\hat{k}$ in $\vec{a}$ is $0$, so they are equal. The length is $\sqrt{2}$, so the direction cosines are $\frac{1}{\sqrt{2}}$, $\frac{1}{\sqrt{2}}$, $0$ (angles $45^{\circ}$, $45^{\circ}$, $90^{\circ}$).

A boat points straight east and its engine gives it a speed of $3$ metres per second through still water. At the same time a river current pushes it north at $4$ metres per second. Where does the boat actually go? Its real velocity is the sum of the two velocities. This is vector addition.

Put the second arrow so that it starts where the first one ends. The sum is the arrow from the very start to the very end. If $\vec{a} = \overrightarrow{AB}$ and $\vec{b} = \overrightarrow{BC}$, then

$$\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}.$$

Here $\vec{a} + \vec{b} = 3\hat{i} + 4\hat{j}$, with magnitude $\sqrt{9 + 16} = 5$. The boat moves at $5$ m/s, in a direction between east and north. Notice that $5 < 3 + 4$: the resultant is shorter than the two sides added, because they do not point the same way.

If two vectors start at the same point, draw the parallelogram they span. The sum is the diagonal from the common starting point. This is the same as the triangle law, with $\vec{b}$ moved to the far side.

In practice, add component by component. For $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + 4\hat{k}$:

$$\vec{a} + \vec{b} = 3\hat{i} + \hat{j} + 3\hat{k}.$$

Subtraction means adding the negative: $\vec{a} - \vec{b} = \vec{a} + (-\vec{b}) = \hat{i} + 5\hat{j} - 5\hat{k}$.

Any side of a triangle is shorter than the other two together, so for all vectors $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$. For the numbers above, $|\vec{a}| \approx 3.742$, $|\vec{b}| \approx 4.583$ and $|\vec{a} + \vec{b}| \approx 4.359$, and indeed $4.359 < 8.324$. Equality happens only when both vectors point the same way.

Vector addition follows the same rules as adding numbers. Each rule was checked with numbers by computer, using $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 4\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{c} = -2\hat{i} + 5\hat{j}$.

• Commutative: $\vec{a} + \vec{b} = \vec{b} + \vec{a}$. Both give $5\hat{i} + \hat{j} + 5\hat{k}$. The order does not matter: the two routes of the parallelogram end at the same corner.
• Associative: $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$. Both give $3\hat{i} + 6\hat{j} + 5\hat{k}$. You can group in any way, so we may write $\vec{a} + \vec{b} + \vec{c}$.
• Zero vector: $\vec{a} + \vec{0} = \vec{a}$. Adding the zero vector changes nothing.
• Negative: $\vec{a} + (-\vec{a}) = \vec{0}$. Going somewhere and coming straight back means no displacement.

Closed polygon. If you walk around a closed shape (a triangle, a square, any polygon) and come back to the start, the displacements add to the zero vector. For the triangle $A$, $B$, $C$: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec{0}$.

Multiplying a vector $\vec{a}$ by a number $\lambda$ stretches or shrinks it. The result $\lambda\vec{a}$ has:

• length $|\lambda\vec{a}| = |\lambda|\,|\vec{a}|$;
• the same direction as $\vec{a}$ if $\lambda > 0$, the opposite direction if $\lambda < 0$;
• the zero vector if $\lambda = 0$.

In components, multiply every component: $\lambda(x\hat{i} + y\hat{j} + z\hat{k}) = \lambda x\,\hat{i} + \lambda y\,\hat{j} + \lambda z\,\hat{k}$.

The unit vector $\hat{a} = \dfrac{\vec{a}}{|\vec{a}|}$ has length $1$. So $m\,\hat{a}$ has length $m$ and the direction of $\vec{a}$. Example: find a vector of length $10$ in the direction of $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$.

The length of $\vec{a}$ is $\sqrt{4 + 9 + 36} = 7$. So the vector is $\dfrac{10}{7}\vec{a} = \dfrac{1}{7}\left(20\hat{i} - 30\hat{j} + 60\hat{k}\right)$. Check: $10\vec{a} = 20\hat{i} - 30\hat{j} + 60\hat{k}$ has length $70$, and dividing by $7$ gives length $10$.

So the answer is $\dfrac{20}{7}\hat{i} - \dfrac{30}{7}\hat{j} + \dfrac{60}{7}\hat{k}$.

Every vector can be written in one way as $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$. The numbers $a_1$, $a_2$, $a_3$ are its components. This is the same as the position vector of the point $(a_1, a_2, a_3)$. Working with components turns geometry into arithmetic.

• Magnitude: $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.
• Equal vectors: $\vec{a} = \vec{b}$ exactly when all three components match.
• Sum and difference: add or subtract the matching components.
• Scalar multiple: multiply each component.
• Collinear (parallel) vectors: $\vec{a} \parallel \vec{b}$ exactly when $\vec{b} = \lambda\vec{a}$ for some number $\lambda$, that is, the components are proportional: $\dfrac{b_1}{a_1} = \dfrac{b_2}{a_2} = \dfrac{b_3}{a_3}$.

A drone climbs with velocity $3\hat{i} + 4\hat{j} + 12\hat{k}$ metres per second ($3$ east, $4$ north, $12$ up). Its speed is $\sqrt{9 + 16 + 144} = 13$ m/s.

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$. Then

• $\vec{a} + \vec{b} = 3\hat{i} - \hat{j} + 4\hat{k}$, with $|\vec{a} + \vec{b}| = \sqrt{26}$.
• $2\vec{a} - 3\vec{b} = 2\hat{i} + 4\hat{j} + 6\hat{k} - \left(6\hat{i} - 9\hat{j} + 3\hat{k}\right) = -4\hat{i} + 13\hat{j} + 3\hat{k}$, with $|2\vec{a} - 3\vec{b}| = \sqrt{194}$.

For which $x$ is $\vec{p} = 2\hat{i} - 4\hat{j} + 6\hat{k}$ parallel to $\vec{q} = x\hat{i} + 2\hat{j} - 3\hat{k}$? The ratio of the $j$-components is $\frac{2}{-4} = -\frac{1}{2}$, and for the $k$-components $\frac{-3}{6} = -\frac{1}{2}$. They agree, so $x = -\frac{1}{2} \times 2 = -1$. The computer confirms $\vec{p}$ and $-\hat{i} + 2\hat{j} - 3\hat{k}$ are parallel.

Write $5\hat{i} + 3\hat{j}$ as $p\,(\hat{i} + 2\hat{j}) + q\,(3\hat{i} - \hat{j})$. Comparing components: $p + 3q = 5$ and $2p - q = 3$. Solving gives $p = 2$ and $q = 1$. Check: $2(\hat{i} + 2\hat{j}) + (3\hat{i} - \hat{j}) = 5\hat{i} + 3\hat{j}$.

Suppose the points $P$ and $Q$ have position vectors $\vec{p} = \overrightarrow{OP}$ and $\vec{q} = \overrightarrow{OQ}$. To walk from $P$ to $Q$ you can go from $P$ back to $O$ and then on to $Q$. In the triangle $OPQ$ the triangle law gives $\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ}$, so

$$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \vec{q} - \vec{p}.$$

Rule: the vector from $P$ to $Q$ is "end minus start", component by component.

Example 1. For $P = (1, 2, 3)$ and $Q = (4, -2, 5)$: $\overrightarrow{PQ} = (4 - 1)\hat{i} + (-2 - 2)\hat{j} + (5 - 3)\hat{k} = 3\hat{i} - 4\hat{j} + 2\hat{k}$, and $|\overrightarrow{PQ}| = \sqrt{9 + 16 + 4} = \sqrt{29}$. The vector $\overrightarrow{QP}$ is the negative: $-3\hat{i} + 4\hat{j} - 2\hat{k}$.

Example 2: a triangle. Take $A = (1, 2, 3)$, $B = (4, 6, 3)$, $C = (2, -1, 5)$. Then $\overrightarrow{AB} = 3\hat{i} + 4\hat{j}$, $\overrightarrow{BC} = -2\hat{i} - 7\hat{j} + 2\hat{k}$ and $\overrightarrow{CA} = -\hat{i} + 3\hat{j} - 2\hat{k}$. Adding the three gives $\vec{0}$, as it must for a closed triangle.

A point $R$ lies on the line through $A$ and $B$ and divides the segment $AB$ in the ratio $m : n$. This means $AR : RB = m : n$. If $A$ and $B$ have position vectors $\vec{a}$ and $\vec{b}$, then $R$ has position vector

$$\vec{r} = \frac{n\,\vec{a} + m\,\vec{b}}{m + n} \qquad (\text{internal division}).$$

Notice the crossing: the far vector gets the larger share when its number $m$ is bigger. The mid-point is the special case $m = n$: $\vec{r} = \dfrac{\vec{a} + \vec{b}}{2}$.

External division. If $R$ lies on the line outside the segment, with $AR : RB = m : n$ (and $m \ne n$), then

$$\vec{r} = \frac{m\,\vec{b} - n\,\vec{a}}{m - n}.$$

Let $A = (1, 2, 3)$ and $B = (6, -3, 8)$. The point dividing $AB$ internally in the ratio $2 : 3$ is $\dfrac{3\vec{a} + 2\vec{b}}{5} = \dfrac{15\hat{i} + 25\hat{k}}{5} = (3, 0, 5)$. Externally in the ratio $3 : 2$ it is $\dfrac{3\vec{b} - 2\vec{a}}{1} = (16, -13, 18)$. The mid-point is $(\frac{7}{2}, -\frac{1}{2}, \frac{11}{2})$.

The centroid $G$ of a triangle is where the three medians meet. It lies $\frac{2}{3}$ of the way from each vertex to the opposite mid-point. For $A = (1, 2, 3)$, $B = (4, 6, 3)$, $C = (7, 1, 0)$: the mid-point of $BC$ is $M = (\frac{11}{2}, \frac{7}{2}, \frac{3}{2})$. Doubling $M$ gives $\vec{b} + \vec{c} = 11\hat{i} + 7\hat{j} + 3\hat{k}$. Dividing $AM$ internally in the ratio $2 : 1$ gives $G = \dfrac{\vec{a} + 2\vec{m}}{3} = \dfrac{\vec{a} + \vec{b} + \vec{c}}{3} = \dfrac{12\hat{i} + 9\hat{j} + 6\hat{k}}{3} = (4, 3, 2)$. In general, the centroid has position vector $\dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$.

Take $A = (1, 2, 3)$, $B = (3, 5, 7)$, $C = (5, 8, 11)$. Then $\overrightarrow{AB} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\overrightarrow{AC} = 4\hat{i} + 6\hat{j} + 8\hat{k}$. Since $\overrightarrow{AC} = 2\overrightarrow{AB}$, the vectors are parallel and share the point $A$, so $A$, $B$, $C$ are on one line. Indeed $B = (3, 5, 7)$ is the mid-point of $AC$.

1. Find $\overrightarrow{PQ}$ and $|\overrightarrow{PQ}|$ for $P = (2, -1, 3)$ and $Q = (-1, 4, 5)$.
2. Find the mid-point of the line segment joining $A = (2, 4, -6)$ and $B = (8, -2, 2)$.
3. Find the point that divides the segment from $P = (-1, 3, 2)$ to $Q = (4, -2, 7)$ internally in the ratio $2 : 3$.
4. Find the point that divides the same segment externally in the ratio $2 : 1$.
5. $ABCD$ is a parallelogram with $A = (1, 1, 1)$, $B = (3, 2, 0)$, $C = (6, 5, 2)$. Find $D$.
6. Show that $A = (1, -1, 2)$, $B = (3, 1, 6)$, $C = (7, 5, 14)$ are on one line.
7. For $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 4\hat{j} - 2\hat{k}$, find $2\vec{a} + 3\vec{b}$ and its magnitude.
8. Write $5\hat{i} + 3\hat{j}$ as $p(\hat{i} + 2\hat{j}) + q(3\hat{i} - \hat{j})$.
9. In what ratio does the plane $y = 0$ (the $xz$-plane) divide the segment joining $A = (1, 2, 3)$ and $B = (6, -3, 8)$? Where does it meet the segment?
10. Find the centroid of the triangle with vertices $(1, 2, 3)$, $(4, 6, 3)$, $(7, 1, 0)$.

Answers

1. $\overrightarrow{PQ} = -3\hat{i} + 5\hat{j} + 2\hat{k}$ and $|\overrightarrow{PQ}| = \sqrt{9 + 25 + 4} = \sqrt{38}$.
2. $\dfrac{\vec{a} + \vec{b}}{2} = \dfrac{10\hat{i} + 2\hat{j} - 4\hat{k}}{2} = (5, 1, -2)$.
3. $\dfrac{3\vec{p} + 2\vec{q}}{5} = \dfrac{5\hat{i} + 5\hat{j} + 20\hat{k}}{5} = (1, 1, 4)$.
4. $\dfrac{2\vec{q} - \vec{p}}{2 - 1} = 9\hat{i} - 7\hat{j} + 12\hat{k} = (9, -7, 12)$.
5. The diagonals of a parallelogram bisect each other, so $A + C = B + D$, so $D = A + C - B = (4, 4, 3)$. Check: the mid-point of $AC$ and the mid-point of $BD$ are both $(\frac{7}{2}, 3, \frac{3}{2})$.
6. $\overrightarrow{AB} = 2\hat{i} + 2\hat{j} + 4\hat{k}$ and $\overrightarrow{AC} = 6\hat{i} + 6\hat{j} + 12\hat{k} = 3\,\overrightarrow{AB}$. The vectors are parallel and share $A$, so the three points are collinear.
7. $2\vec{a} + 3\vec{b} = 7\hat{i} + 10\hat{j}$, with magnitude $\sqrt{49 + 100} = \sqrt{149}$.
8. $p + 3q = 5$ and $2p - q = 3$ give $p = 2$, $q = 1$. Check: $2(\hat{i} + 2\hat{j}) + (3\hat{i} - \hat{j}) = 5\hat{i} + 3\hat{j}$.
9. Let the ratio be $m : n$. The $y$-coordinate is $\dfrac{n \cdot 2 + m \cdot (-3)}{m + n} = 0$, so $2n = 3m$ and $m : n = 2 : 3$. The meeting point is $(3, 0, 5)$.
10. $\dfrac{\vec{a} + \vec{b} + \vec{c}}{3} = \dfrac{12\hat{i} + 9\hat{j} + 6\hat{k}}{3} = (4, 3, 2)$.

You can add two vectors and you can scale one. Can you multiply two vectors? There is more than one sensible answer, because two arrows can be combined in two useful ways:

• The scalar product (or dot product), written $\vec{a} \cdot \vec{b}$. The answer is a number. It measures how much the two vectors point the same way. Example: work $= $ force $\cdot$ displacement.
• The vector product (or cross product), written $\vec{a} \times \vec{b}$. The answer is a vector that is perpendicular to both. It measures how much they point different ways. Examples: the area of a parallelogram, and the turning effect (torque) of a force.

Let $\theta$ be the angle between two non-zero vectors $\vec{a}$ and $\vec{b}$ when they are drawn from the same point, with $0 \le \theta \le \pi$. The scalar product is

$$\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta.$$

If either vector is the zero vector, the product is $0$. The answer is a plain number, which can be positive, zero or negative, depending on whether $\theta$ is less than, equal to, or more than $90^{\circ}$.

A force $\vec{F}$ moves an object through a displacement $\vec{d}$. Only the part of the force along the motion does work. So the work done is $W = |\vec{F}|\,|\vec{d}|\cos\theta = \vec{F} \cdot \vec{d}$. For example, a force $\vec{F} = 6\hat{i} + 8\hat{j}$ newtons moves a box by $\vec{d} = 5\hat{i}$ metres. The work is $30$ joules: the $8$ in the $\hat{j}$-direction is sideways to the motion and contributes nothing.

Writing both vectors in components, the dot product is simply

$$\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3.$$

Example. For $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 4\hat{i} - \hat{j} + 2\hat{k}$: $\vec{a} \cdot \vec{b} = 1 \cdot 4 + 2 \cdot (-1) + 3 \cdot 2 = 8$. The lengths are $|\vec{a}| = \sqrt{14}$ and $|\vec{b}| = \sqrt{21}$. So

$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|} = \frac{8}{\sqrt{294}}$$

So $\theta \approx 62.19^{\circ}$.

• Perpendicular vectors. If $\vec{a}$ and $\vec{b}$ are non-zero, then $\vec{a} \cdot \vec{b} = 0$ exactly when they are perpendicular (because $\cos 90^{\circ} = 0$). Example: $\hat{i} + 2\hat{j} + 2\hat{k}$ and $2\hat{i} + \hat{j} - 2\hat{k}$ give $1 \cdot 2 + 2 \cdot 1 + 2 \cdot (-2) = 0$, so they are perpendicular.
• The sign tells the angle. Positive means $\theta < 90^{\circ}$ (acute). Zero means $90^{\circ}$. Negative means $\theta > 90^{\circ}$ (obtuse). Example: $\hat{i}$ and $-\hat{i} + \hat{j}$ give $-1$, and the angle is $135^{\circ}$.
• A vector with itself. $\vec{a} \cdot \vec{a} = |\vec{a}|^2$, since $\theta = 0$ and $\cos 0 = 1$. For $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ this is $1 + 4 + 9 = 14$. This is how we get the length: $|\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}$.
• Order does not matter. $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$.
• The unit vectors. Each of $\hat{i}$, $\hat{j}$, $\hat{k}$ has length $1$, and the three are mutually perpendicular. So: