Exercise 3.3 — Question 5

Cross Product (Vector Product)

The cross product (or vector product) of two vectors $\vec{a}$ and $\vec{b}$ is a vector $\vec{c}$ perpendicular to both $\vec{a}$ and $\vec{b}$, defined as:

$\vec{a}\times \vec{b}=|\vec{a}||\vec{b}|\sin \theta \hat{n}$

where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, and $\hat{n}$ is the unit vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ (direction given by the right-hand rule).

Component Form

If $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, then:

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

$=\hat{i}(a_2{b}_3-a_3{b}_2)-\hat{j}(a_1{b}_3-a_3{b}_1)+\hat{k}(a_1{b}_2-a_2{b}_1)$

Finding the Angle Using Cross Product

The magnitude of the cross product gives:

$|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta$

Solving for $\theta$:

$\sin \theta =\frac{|\vec{a}\times \vec{b}|}{|\vec{a}||\vec{b}|}$

$\theta ={\sin }^{-1}\left(\frac{|\vec{a}\times \vec{b}|}{|\vec{a}||\vec{b}|}\right)$

Geometrical Interpretation

The magnitude $|\vec{a}\times \vec{b}|$ equals the area of the parallelogram formed by vectors $\vec{a}$ and $\vec{b}$ as adjacent sides.

Key Properties

• $\vec{a}\times \vec{b}=-\vec{b}\times \vec{a}$ (anti-commutative)
• $\vec{a}\times \vec{a}=\vec{0}$ (cross product of a vector with itself is zero)
• $\hat{i}\times \hat{j}=\hat{k},\hat{j}\times \hat{k}=\hat{i},\hat{k}\times \hat{i}=\hat{j}$
• $\hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=\vec{0}$
• Two non-zero vectors $\vec{a}$ and $\vec{b}$ are parallel if and only if $\vec{a}\times \vec{b}=\vec{0}$

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Worked Example

Given: $\vec{a}=2\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$

Find: (i) $\vec{a}\times \vec{b}$, (ii) the angle $\theta$ between $\vec{a}$ and $\vec{b}$

Step 1: Compute the cross product

$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 1& -1& 2\end{array}\right|$

$=\hat{i}[(1)(2)-(-1)(-1)]-\hat{j}[(2)(2)-(-1)(1)]+\hat{k}[(2)(-1)-(1)(1)]$

$=\hat{i}[2-1]-\hat{j}[4+1]+\hat{k}[-2-1]$

$=\hat{i}-5\hat{j}-3\hat{k}$

Step 2: Find magnitudes

$|\vec{a}\times \vec{b}|=\sqrt{1^2+(-5{)}^2+(-3{)}^2}=\sqrt{1+25+9}=\sqrt{35}$

$|\vec{a}|=\sqrt{4+1+1}=\sqrt{6},|\vec{b}|=\sqrt{1+1+4}=\sqrt{6}$

Step 3: Find the angle

$\sin \theta =\frac{\sqrt{35}}{\sqrt{6}\cdot \sqrt{6}}=\frac{\sqrt{35}}{6}$

$\theta ={\sin }^{-1}\left(\frac{\sqrt{35}}{6}\right)$