3.3 Q-7

Exercise 3.3 — Question 7

Problem

Find the cross product $\vec{a}\times \vec{b}$ for the given vectors, and use it to find the angle between them.

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Key Concepts

Cross Product Definition

The cross product (vector product) of two vectors $\vec{a}$ and $\vec{b}$ is defined as:

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

where $\theta$ is the angle between the vectors and $\hat{n}$ is the unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ (determined by the right-hand rule).

Cross Product in Component Form

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

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

Expanding:

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

Finding the Angle Using Cross Product

The magnitude of the cross product gives:

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

Therefore:

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

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

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Worked Example

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

Step 1: Compute the cross product

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

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

$=\hat{i}(2-1)-\hat{j}(4+1)+\hat{k}(-2-1)$

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

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

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

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

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Geometrical Interpretation

• $|\vec{a}\times \vec{b}|$ equals the area of the parallelogram formed by $\vec{a}$ and $\vec{b}$.
• The direction of $\vec{a}\times \vec{b}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$.
• If $\vec{a}\times \vec{b}=\vec{0}$, the vectors are parallel (or one is the zero vector).