Vectors and the dot product

In the second half of week 6 the subject of the course changes completely: we move from series to the geometry of three-dimensional space. Every topic from here on (lines and planes, surfaces, gradient, directional derivative, tangent plane, Lagrange multipliers) is written in the language of vectors. So being able to do the operations on this page quickly and without mistakes will make the next eight weeks easier. The exam has direct “find the angle / projection / work” questions, but their real importance is that they are used constantly as intermediate steps in later topics.

Golden rules:

1. Vector = direction + length. Its components are written $\langle v_1,v_2,v_3\rangle$. $\overrightarrow{PQ}$ = end minus start.
2. The result of the dot product is a number, not a vector.
3. If $\mathbf u\cdot\mathbf v = 0$, the vectors are perpendicular. This is the one-line test for all perpendicularity questions.
4. In the projection formula the denominator has $\lvert\mathbf v\rvert^2$ (vector projection), and in the scalar component it has $\lvert\mathbf v\rvert$. Do not mix them up.

Vectors: definition and basic operations

In space we denote a point by $(x,y,z)$ and a vector by $\langle v_1,v_2,v_3\rangle$. The starting point of a vector does not matter; all arrows with the same direction and length are the same vector. The vector from $P(x_1,y_1,z_1)$ to $Q(x_2,y_2,z_2)$ is

$\overrightarrow{PQ} = \langle x_2-x_1,\;y_2-y_1,\;z_2-z_1\rangle.$

It is also written with the standard unit vectors as $\mathbf v = v_1\mathbf i+v_2\mathbf j+v_3\mathbf k$; $\mathbf i = \langle1,0,0\rangle$, $\mathbf j = \langle0,1,0\rangle$, $\mathbf k = \langle0,0,1\rangle$.

The length of a unit vector is $1$; it carries only direction. If “a vector of length $5$ in the direction of $\mathbf v$” is asked, you write $5\dfrac{\mathbf v}{\lvert\mathbf v\rvert}$.

Intuition. Addition is “placing end to end”: first go $\mathbf u$, then from there go $\mathbf v$; the place you reach is $\mathbf u+\mathbf v$. The parallelogram rule says the same thing.

The dot product

Definition.

$\mathbf u\cdot\mathbf v = u_1v_1+u_2v_2+u_3v_3.$

Geometric meaning. If $\theta$ is the angle between the two vectors ($0\le\theta\le\pi$), then

$\mathbf u\cdot\mathbf v = \lvert\mathbf u\rvert\,\lvert\mathbf v\rvert\cos\theta\qquad\Longrightarrow\qquad\cos\theta = \frac{\mathbf u\cdot\mathbf v}{\lvert\mathbf u\rvert\,\lvert\mathbf v\rvert}.$

Properties. $\mathbf u\cdot\mathbf v = \mathbf v\cdot\mathbf u$; $\mathbf u\cdot(\mathbf v+\mathbf w) = \mathbf u\cdot\mathbf v+\mathbf u\cdot\mathbf w$; $\mathbf v\cdot\mathbf v = \lvert\mathbf v\rvert^2$; $\mathbf 0\cdot\mathbf v = 0$.

Intuition. The dot product answers the question “how much do two vectors face the same direction”. If they point the same way, $\cos\theta = 1$ and the product is largest. If they are perpendicular, it is $0$. If they point opposite ways, it is negative.

Example 1. The angle between $\mathbf u = \langle2,-1,3\rangle$ and $\mathbf v = \langle1,4,-2\rangle$.

u·v   = 2·1 + (-1)·4 + 3·(-2) = 2 - 4 - 6 = -8
|u|   = √(4+1+9)  = √14
|v|   = √(1+16+4) = √21
cos θ = -8/√294 ≈ -0.4666
θ     ≈ 117.81°     (negative product → obtuse angle, consistent)

Example 2 (perpendicularity). $\langle1,2,3\rangle$ and $\langle3,0,-1\rangle$: $3+0-3 = 0$, they are perpendicular.

PITFALL · the range of the angle. $\arccos$ always gives values in $[0,\pi]$. The angle between vectors cannot be negative or greater than $180^\circ$. Use the calculator in the right mode (degrees/radians) and see which one the question wants.

Projection

The projection of $\mathbf u$ onto $\mathbf v$ is the shadow formed when you drop a perpendicular from the tip of $\mathbf u$ onto the line of $\mathbf v$.

$\text{proj}_{\mathbf v}\mathbf u = \left(\frac{\mathbf u\cdot\mathbf v}{\lvert\mathbf v\rvert^2}\right)\mathbf v\qquad\text{(vector)}$

$\text{comp}_{\mathbf v}\mathbf u = \frac{\mathbf u\cdot\mathbf v}{\lvert\mathbf v\rvert} = \lvert\mathbf u\rvert\cos\theta\qquad\text{(scalar, signed length)}$

The projection is parallel to $\mathbf v$. And $\mathbf u-\text{proj}_{\mathbf v}\mathbf u$ is perpendicular to $\mathbf v$. This way you can split any vector into two parts, “parallel + perpendicular”. In physics, splitting a force into its components parallel and perpendicular to an inclined plane is exactly this.

Example 3. With the vectors of Example 1: $\text{proj}_{\mathbf v}\mathbf u = \dfrac{-8}{21}\langle1,4,-2\rangle$. The coefficient is negative, so the projection points opposite to $\mathbf v$. This is consistent with the angle being obtuse.

TIP · onto which one? In $\text{proj}_{\mathbf v}\mathbf u$ the denominator has the length of the vector onto which you project ($\mathbf v$) and the result is in its direction. If you swap $\mathbf u$ and $\mathbf v$ you get a completely different vector.

Direction angles and direction cosines

The angles $\alpha,\beta,\gamma$ that $\mathbf v$ makes with the $x$, $y$, $z$ axes:

$\cos\alpha = \frac{v_1}{\lvert\mathbf v\rvert},\quad\cos\beta = \frac{v_2}{\lvert\mathbf v\rvert},\quad\cos\gamma = \frac{v_3}{\lvert\mathbf v\rvert},\qquad\cos^2\alpha+\cos^2\beta+\cos^2\gamma = 1.$

The direction cosines are really the components of the unit vector. That is why the sum of their squares is $1$.

Work

If a constant force $\mathbf F$ moves an object by $\mathbf D$, the work done is

$W = \mathbf F\cdot\mathbf D = \lvert\mathbf F\rvert\,\lvert\mathbf D\rvert\cos\theta.$

Only the component of the force in the direction of motion does work. If the force is perpendicular to the direction of motion, the work is zero.

Example 4. $\lvert\mathbf F\rvert = 5$ N, $\mathbf F = \langle3,0,4\rangle$ and the object is displaced by $\langle6,0,8\rangle$ m: $W = 18+32 = 50$ J. The force and the displacement are in the same direction ($\mathbf D = 2\mathbf F$), $\cos\theta = 1$, $W = 5\cdot10 = 50$. The same result by two routes.

Example 5 (with an angle given). A force of $40$ N pulls a box $10$ m horizontally with a rope that makes a $60^\circ$ angle with the horizontal: $W = 40\cdot10\cdot\cos60^\circ = 200$ J.

Common mistakes