Vectors: Cross Product (Exercise 10.4, Part 1)

This lesson works through five cross product questions from Exercise 10.4 for Class XII. Each one expands a determinant, takes a magnitude, or uses those results to find a perpendicular vector or an angle.

Magnitude of a cross product

Given $\vec{a} = \mathbf{i} - 7\mathbf{j} + 7\mathbf{k}$ and $\vec{b} = 3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$, find $|\vec{a} \times \vec{b}|$.

Set up the determinant with the unit vectors in the first row and the components of $\vec{a}$ and $\vec{b}$ below:

$\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix}$

Expanding along the first row:

$\vec{a} \times \vec{b} = \mathbf{i}\big((-7)(2) - (7)(-2)\big) - \mathbf{j}\big((1)(2) - (7)(3)\big) + \mathbf{k}\big((1)(-2) - (-7)(3)\big)$

$= \mathbf{i}(0) - \mathbf{j}(-19) + \mathbf{k}(19) = 19\mathbf{j} + 19\mathbf{k}$

The magnitude is the square root of the sum of the squares of the components:

$|\vec{a} \times \vec{b}| = \sqrt{19^2 + 19^2} = 19\sqrt{2}$

Unit vector perpendicular to two vectors

Find a unit vector perpendicular to both $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$, where $\vec{a} = 3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ and $\vec{b} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$.

Set up the two vectors. Let

$\vec{c} = \vec{a} + \vec{b} = 4\mathbf{i} + 4\mathbf{j} + 0\mathbf{k}, \qquad \vec{d} = \vec{a} - \vec{b} = 2\mathbf{i} + 0\mathbf{j} + 4\mathbf{k}$

Cross product. A vector perpendicular to both is $\vec{c} \times \vec{d}$:

$\vec{c} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = \mathbf{i}(16 - 0) - \mathbf{j}(16 - 0) + \mathbf{k}(0 - 8) = 16\mathbf{i} - 16\mathbf{j} - 8\mathbf{k}$

Magnitude and unit vector.

$|\vec{c} \times \vec{d}| = \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{576} = 24$

Taking out a factor of $8$ from the numerator and dividing by $24$:

$\hat{n} = \pm \frac{16\mathbf{i} - 16\mathbf{j} - 8\mathbf{k}}{24} = \pm \frac{2\mathbf{i} - 2\mathbf{j} - \mathbf{k}}{3}$

The two perpendicular unit vectors are $\tfrac{1}{3}(2\mathbf{i} - 2\mathbf{j} - \mathbf{k})$ and $\tfrac{1}{3}(-2\mathbf{i} + 2\mathbf{j} + \mathbf{k})$.

Finding an angle from direction cosines

A unit vector makes an angle $\tfrac{\pi}{3}$ with $\mathbf{i}$, $\tfrac{\pi}{4}$ with $\mathbf{j}$, and an acute angle $\theta$ with $\mathbf{k}$. Find $\theta$ and the components of $\vec{a}$.

The direction cosines are $l = \cos\tfrac{\pi}{3}$, $m = \cos\tfrac{\pi}{4}$, $n = \cos\theta$, so

$\vec{a} = \tfrac{1}{2}\mathbf{i} + \tfrac{1}{\sqrt{2}}\mathbf{j} + \cos\theta\,\mathbf{k}$

Since $\vec{a}$ is a unit vector, $|\vec{a}| = 1$, and squaring gives

$\left(\tfrac{1}{2}\right)^2 + \left(\tfrac{1}{\sqrt{2}}\right)^2 + \cos^2\theta = 1$

$\tfrac{1}{4} + \tfrac{1}{2} + \cos^2\theta = 1 \implies \cos^2\theta = 1 - \tfrac{3}{4} = \tfrac{1}{4}$

So $\cos\theta = \pm\tfrac{1}{2}$. As $\theta$ is acute, $\cos\theta = \tfrac{1}{2}$ and $\theta = \tfrac{\pi}{3}$. The components of $\vec{a}$ are $\tfrac{1}{2}$, $\tfrac{1}{\sqrt{2}}$, and $\tfrac{1}{2}$.

Solving for unknowns that make a cross product zero

Find $\lambda$ and $\mu$ if $(2\mathbf{i} + 6\mathbf{j} + 27\mathbf{k}) \times (\mathbf{i} + \lambda\mathbf{j} + \mu\mathbf{k}) = \vec{0}$.

The cross product is zero, so expand the determinant and set each component to zero:

$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} = \mathbf{i}(6\mu - 27\lambda) - \mathbf{j}(2\mu - 27) + \mathbf{k}(2\lambda - 6) = \vec{0}$

This gives three equations:

$6\mu - 27\lambda = 0 \quad (1), \qquad 2\mu - 27 = 0 \quad (2), \qquad 2\lambda - 6 = 0 \quad (3)$

From $(2)$, $\mu = \tfrac{27}{2}$. From $(3)$, $\lambda = 3$.

Check in (1). $6 \cdot \tfrac{27}{2} - 27 \cdot 3 = 81 - 81 = 0$, so the values satisfy the first equation. Hence $\lambda = 3$ and $\mu = \tfrac{27}{2}$.

Angle between two vectors from their cross product

Let $\vec{a}$ and $\vec{b}$ satisfy $|\vec{a}| = 3$ and $|\vec{b}| = \tfrac{\sqrt{2}}{3}$, and let $\vec{a} \times \vec{b}$ be a unit vector. Find the angle between them.

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

$\sin\theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}|\,|\vec{b}|} = \frac{1}{3 \cdot \tfrac{\sqrt{2}}{3}} = \frac{1}{\sqrt{2}}$

So $\theta = \tfrac{\pi}{4}$.

Key takeaways

• A cross product is found by expanding the determinant with the unit vectors in the top row, and its magnitude is the root of the sum of the squares of the components.
• $\vec{c} \times \vec{d}$ is perpendicular to both $\vec{c}$ and $\vec{d}$; dividing it by its magnitude gives the two perpendicular unit vectors $\pm\hat{n}$.
• For a unit vector, the squares of its direction cosines add to $1$, which lets you solve for an unknown angle.
• If a cross product is the zero vector the vectors are parallel; if it is a unit vector, $\sin\theta = \dfrac{1}{|\vec{a}|\,|\vec{b}|}$ gives the angle between them.