Vector Space ℝ³ of Real Numbers

Lessons

Senior 5 · Mathematics

Vector Space ℝ³ of Real Numbers

4 lessons

Lesson 3 of 4

Scalar product, angles and direction cosines

40 min

How much of a force does the work?

A farmer pulls a loaded wooden cart up a sloping path with a rope. The rope is not parallel to the path, so only part of the pull moves the cart forward; the rest just tugs it towards the bank. Physics measures the useful part as work, and computing it needs one operation on two vectors that returns a number: the scalar product. The same product gives lengths, angles, perpendicular directions and projections in $\mathbb{R}^3$ . In S4 you used it in the plane; here it gains a third component.

Definition

The scalar product (dot product) of $u = (u_{1}, u_{2}, u_{3})$ and $v = (v_{1}, v_{2}, v_{3})$ is the real number

u \cdot v = u_1v_1 + u_2v_2 + u_3v_3 .

The magnitude (norm, length) of $u$ is $|u| = \sqrt{u \cdot u} = \sqrt{u_1^2 + u_2^2 + u_3^2}$ . A unit vector has magnitude 1, and $\hat{u} = \frac{u}{|u|}$ is the unit vector in the direction of $u \ne 0$ . The distance between points $A$ and $B$ is $AB = |\overrightarrow{AB}|$ .

Example: $(2, -1, 2) \cdot (3, 0, 4) = 6 + 0 + 8 = 14$ , and $|(2, -1, 2)| = \sqrt{4 + 1 + 4} = 3$ .

Important

Properties. For $u, v, w \in \mathbb{R}^3$ and $\lambda \in \mathbb{R}$ : $u \cdot v = v \cdot u$ ; $u \cdot (v + w) = u \cdot v + u \cdot w$ ; $(\lambda u) \cdot v = \lambda(u \cdot v)$ ; $u \cdot u \ge 0$ , with $u \cdot u = 0$ only for $u = 0$ .

Theorem (angle form). If $\theta \in [0^\circ, 180^\circ]$ is the angle between non-zero vectors $u$ and $v$ , then

u \cdot v = |u|\,|v|\cos\theta, \qquad \text{so} \qquad \cos\theta = \frac{u \cdot v}{|u|\,|v|} .

Reason. Draw $u$ and $v$ from one point; the third side of the triangle is $u - v$ . The cosine rule gives $|u - v|^2 = |u|^2 + |v|^2 - 2|u||v|\cos\theta$ . Expanding with the properties, $|u - v|^2 = (u - v) \cdot (u - v) = |u|^2 - 2u \cdot v + |v|^2$ . Comparing the two gives the formula.

Consequence. Non-zero vectors are orthogonal (perpendicular) if and only if $u \cdot v = 0$ . The angle is acute when $u \cdot v > 0$ and obtuse when $u \cdot v < 0$ .

Worked example 1: the angle between two vectors

Find the angle between $u = (2, - 1, 2)$ and $v = (3, 0, 4)$ .

Solution. $u \cdot v = 14$ , $∣ u ∣ = 3$ , $|v| = \sqrt{9 + 0 + 16} = 5$ . So $\cos\theta = \frac{14}{15}$ and $\theta = \cos^{-1}\frac{14}{15} \approx 21.0^\circ$ .

Worked example 2: an unknown component for a right angle

Find $k$ so that $(k, 2, - 1)$ and $(3, k, 5)$ are orthogonal.

Solution. $(k, 2, -1) \cdot (3, k, 5) = 3k + 2k - 5 = 5k - 5 = 0$ , so $k = 1$ . Check: $(1, 2, -1) \cdot (3, 1, 5) = 3 + 2 - 5 = 0$ . ✓

Definition

The scalar projection of $u$ on a non-zero vector $v$ is $\frac{u \cdot v}{|v|} = |u|\cos\theta$ , the signed length of the "shadow" of $u$ along $v$ . The vector projection is that length times $\hat{v}$ :

\operatorname{proj}_v u = \frac{u \cdot v}{|v|^2}\, v .

Example: for $u = (4, 1, - 2)$ and $v = (2, - 1, 2)$ , $u \cdot v = 8 - 1 - 4 = 3$ and $∣ v ∣ = 3$ , so the scalar projection is $1$ and $\operatorname{proj}_v u = \frac{3}{9}(2, -1, 2) = \left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right)$ .

Definition

The direction angles of a non-zero vector $u$ are the angles $\alpha, \beta, \gamma$ that it makes with the positive $x$ -, $y$ - and $z$ -axes. Their cosines are the direction cosines of $u$ :

\cos\alpha = \frac{u_1}{|u|}, \qquad \cos\beta = \frac{u_2}{|u|}, \qquad \cos\gamma = \frac{u_3}{|u|} .

(Each follows from the angle formula with $v = i$ , $j$ or $k$ .) So $\hat{u} = (\cos\alpha, \cos\beta, \cos\gamma)$ , and since $|\hat{u}| = 1$ ,

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 .

Worked example 3: direction cosines and angles

Find the direction cosines and direction angles of $u = (2, - 1, 2)$ .

Solution. $∣ u ∣ = 3$ , so $\cos\alpha = \frac{2}{3}$ , $\cos\beta = -\frac{1}{3}$ , $\cos\gamma = \frac{2}{3}$ . Check: $\frac{4}{9} + \frac{1}{9} + \frac{4}{9} = 1$ . ✓ The angles are $\alpha = \gamma = \cos^{-1}\frac{2}{3} \approx 48.2^\circ$ and $\beta = \cos^{-1}\left(-\frac{1}{3}\right) \approx 109.5^\circ$ : $u$ points away from the positive $y$ -axis.

Worked example 4: work done by a constant force

A constant force $F = 3 i + 2 j + 4 k$ (in newtons) moves a crate in a straight line from $A (1, 0, 2)$ to $B (4, 5, 3)$ (in metres). Find the work done and the angle between the force and the displacement.

Solution. The work done by a constant force $F$ over a displacement $d$ is $W = F \cdot d$ . Here $d = \overrightarrow{AB} = (3, 5, 1)$ , so

W = 3(3) + 2(5) + 4(1) = 23\ \mathrm{J} .

With $|F| = \sqrt{29}$ and $|d| = \sqrt{35}$ , $\cos\theta = \frac{23}{\sqrt{29}\sqrt{35}} \approx 0.722$ , so $\theta \approx 43.8^\circ$ . Only the component of $F$ along $d$ does work.

Warning

The scalar product is a number, not a vector. Writing $u \cdot v = (6, 0, 8)$ instead of $14$ is a common slip.
$|u| \ne u_1 + u_2 + u_3$ . Square, add, then take the square root: $|(1, 2, 2)| = \sqrt{1 + 4 + 4} = 3$ , not $1 + 2 + 2 = 5$ .
Set your calculator to degrees (or radians) and say which you used; give angles in $[0^\circ, 180^\circ]$ .

Tip

Perpendicular test in one line: multiply matching components, add, and see if you get 0. No square roots needed.

MIT OpenCourseWare: Lecture 1 of 18.02 Multivariable Calculus, Dot product, with Denis Auroux (about 50 min). The first half covers vectors in space, the dot product, lengths and angles, as in this lesson.

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Practice

Questions 1 to 4 are routine; 5 to 7 need more care; 8 asks for a proof.

For $u = (1, - 2, 2)$ and $v = (4, 0, - 3)$ , find $u \cdot v$ , $∣ u ∣$ , $∣ v ∣$ and the angle between them.
Find the unit vector in the direction of $(6, - 2, 3)$ .
Find the distance between $A (1, - 2, 3)$ and $B (4, 2, 15)$ .
Find $m$ so that $(m, 1, 3)$ and $(2, m, - 1)$ are orthogonal.
A vector makes angles of $60^\circ$ with the $x$ -axis and $45^\circ$ with the $y$ -axis. Find the possible angles with the $z$ -axis.
Show that the triangle with vertices $A (1, 0, 1)$ , $B (2, 1, 3)$ , $C (3, - 1, 2)$ is equilateral, and check that $\angle BAC = 60^\circ$ with the scalar product.
A force $F = (2, - 1, 3)$ N moves an object from $(0, 1, 1)$ to $(3, 4, 2)$ , in metres. Find the work done.
Prove that $∣ u + v ∣^{2} + ∣ u - v ∣^{2} = 2 ∣ u ∣^{2} + 2 ∣ v ∣^{2}$ for all $u, v \in \mathbb{R}^3$ . What does it say about a parallelogram?

Show answersHide answers

$u \cdot v = 4 + 0 - 6 = -2$ ; $∣ u ∣ = 3$ ; $∣ v ∣ = 5$ ; $\cos\theta = -\frac{2}{15}$ , $\theta \approx 97.7^\circ$ .
$∣ (6, - 2, 3) ∣ = 7$ , so $\left(\frac{6}{7}, -\frac{2}{7}, \frac{3}{7}\right)$ .
$\overrightarrow{AB} = (3, 4, 12)$ , $AB = \sqrt{9 + 16 + 144} = 13$ .
$2 m + m - 3 = 0$ , so $m = 1$ .
$\cos^2\gamma = 1 - \frac{1}{4} - \frac{1}{2} = \frac{1}{4}$ , so $\cos\gamma = \pm\frac{1}{2}$ : $\gamma = 60^\circ$ or $120^\circ$ .
$\overrightarrow{AB} = (1, 1, 2)$ , $\overrightarrow{AC} = (2, -1, 1)$ , $\overrightarrow{BC} = (1, -2, -1)$ : all have length $\sqrt{6}$ . $\cos\angle BAC = \frac{2 - 1 + 2}{\sqrt{6}\sqrt{6}} = \frac{1}{2}$ , so $60^\circ$ .
$d = (3, 3, 1)$ , $W = 6 - 3 + 3 = 6\ \mathrm{J}$ .
Expand: $(u + v) \cdot (u + v) + (u - v) \cdot (u - v) = |u|^2 + 2u \cdot v + |v|^2 + |u|^2 - 2u \cdot v + |v|^2$ . In a parallelogram, the sum of the squares of the diagonals equals the sum of the squares of the four sides.

Sources and credits

Written by Ganzaa: the explanations, the diagram of direction angles (drawn from computed coordinates), the worked examples, the practice items and the quiz questions. Every product, length and angle was checked by computer.
Video: Lec 1: Dot product, MIT 18.02 Multivariable Calculus, Fall 2007, Denis Auroux, MIT OpenCourseWare, embedded from YouTube.

1How much of a force does the work?

How much of a force does the work?