How Two Kinds of Vector Multiplication Measure Angles and Areas

Vectors can be combined by multiplication in two useful ways. The dot product gives a number that measures how much two vectors point the same way; the cross product gives a new vector perpendicular to both, whose length measures an area.

This part covers the dot product, angles and projections, the cross product, and areas and parallel vectors.

**The dot product is the scalar $\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta$, which in components equals $a_1{b}_1+a_2{b}_2+a_3{b}_3$; geometrically it is the length of one vector times the projection of the other onto it.

Component form.** For $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$:

$$\vec{a}\cdot \vec{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$$

Properties:

- Commutative — $\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$
- Self-product — $\vec{a}\cdot \vec{a}=|\vec{a}{|}^2$
- Unit vectors — $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1$, while $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0$

Worked example. For $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+4\hat{k}$:

$$\vec{a}\cdot \vec{b}=2(1)+3(-1)+(-1)(4)=-5$$

The negative value shows that the angle between them is obtuse.

An everyday example. Pulling a loaded trolley by a rope at an angle does work equal to the dot product of force and displacement — only the part of the pull along the ground counts.

The substance. The sign of the dot product reveals the angle — positive for acute, zero for a right angle, negative for obtuse.

**The angle between two vectors satisfies $\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$, the vectors are perpendicular exactly when $\vec{a}\cdot \vec{b}=0$, and the projection of $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$.

Worked example 1 — angle.** For $\vec{a}=\hat{i}+\hat{j}$ and $\vec{b}=\hat{j}+\hat{k}$:

$$\cos \theta =\frac{0+1+0}{\sqrt{2}\sqrt{2}}=\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}$$

Worked example 2 — perpendicular. Find $\lambda$ so that $2\hat{i}+\lambda \hat{j}+\hat{k}$ is perpendicular to $\hat{i}-2\hat{j}+3\hat{k}$:

$$2-2\lambda +3=0\Rightarrow \lambda =\frac{5}{2}$$

Worked example 3 — projection. The projection of $\vec{a}=2\hat{i}+3\hat{j}+2\hat{k}$ on $\vec{b}=\hat{i}+2\hat{j}+\hat{k}$ is

$$\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\frac{2+6+2}{\sqrt{6}}=\frac{10}{\sqrt{6}}$$

An everyday example. The shadow of a leaning pole on flat ground when the sun is overhead is its projection onto the ground.

The substance. **The projection of $\vec{a}$ on $\vec{b}$ differs from that of $\vec{b}$ on $\vec{a}$** — the divisor changes from $|\vec{b}|$ to $|\vec{a}|$.

**The cross product $\vec{a}\times \vec{b}$ is a vector of magnitude $|\vec{a}||\vec{b}|\sin \theta$, perpendicular to both $\vec{a}$ and $\vec{b}$ with direction set by the right-hand rule, so reversing the order reverses the direction: $\vec{b}\times \vec{a}=-\vec{a}\times \vec{b}$.

Determinant form:**

$$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$$

Properties:

- Anti-commutative — $\vec{a}\times \vec{b}=-\vec{b}\times \vec{a}$
- Parallel vectors — $\vec{a}\times \vec{b}=\vec{0}$; in particular $\vec{a}\times \vec{a}=\vec{0}$
- Unit vectors — $\hat{i}\times \hat{j}=\hat{k}$, $\hat{j}\times \hat{k}=\hat{i}$, $\hat{k}\times \hat{i}=\hat{j}$

Worked example. For $\vec{a}=2\hat{i}+\hat{j}+3\hat{k}$ and $\vec{b}=3\hat{i}+5\hat{j}-2\hat{k}$:

$$\vec{a}\times \vec{b}=\hat{i}(-2-15)-\hat{j}(-4-9)+\hat{k}(10-3)=-17\hat{i}+13\hat{j}+7\hat{k}$$

Check. $\vec{a}\cdot (\vec{a}\times \vec{b})=-34+13+21=0$, so the result is perpendicular to $\vec{a}$.

An everyday example. Tightening a nut with a spanner produces a turning effect along the bolt, perpendicular to both the spanner and the push — the direction of a cross product.

The substance. Swapping the order of the factors flips the vector, because the right-hand rule depends on the order.

**The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is $|\vec{a}\times \vec{b}|$, a triangle has half that area, and two non-zero vectors are parallel exactly when their cross product is the zero vector.

Worked example 1 — parallelogram.** Adjacent sides $\vec{a}=3\hat{i}+\hat{j}+4\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$:

$$\vec{a}\times \vec{b}=\hat{i}(1+4)-\hat{j}(3-4)+\hat{k}(-3-1)=5\hat{i}+\hat{j}-4\hat{k}$$

$$\text{Area}=\sqrt{25+1+16}=\sqrt{42}\text{ sq units}$$

Worked example 2 — triangle. Vertices $A(1,1,1)$, $B(1,2,3)$ and $C(2,3,1)$ give $\overrightarrow{AB}=\hat{j}+2\hat{k}$ and $\overrightarrow{AC}=\hat{i}+2\hat{j}$:

$$\overrightarrow{AB}\times \overrightarrow{AC}=-4\hat{i}+2\hat{j}-\hat{k}\text{Area}=\frac{1}{2}\sqrt{16+4+1}=\frac{\sqrt{21}}{2}\text{ sq units}$$

Worked example 3 — parallel test. For $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$ and $\vec{b}=-2\hat{i}+4\hat{j}-6\hat{k}$, $\vec{b}=-2\vec{a}$, so $\vec{a}\times \vec{b}=\vec{0}$ and the vectors are parallel.

An everyday example. A triangular section of tiled floor with two known edge vectors has its area given directly by half the cross product, with no height to measure.

The substance. The parallelogram formula uses adjacent sides, not diagonals — with diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$, the area is $\frac{1}{2}|\overrightarrow{d_1}\times \overrightarrow{d_2}|$.