3.3 Q-6

Exercise 3.3 — Question 6

Topic: Cross Product and Angle Between Vectors

This question applies the cross product (vector product) of two vectors to find the angle between them, and explores the geometrical interpretation of the cross product.

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Key Concepts

Cross Product Definition

The cross 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 $\vec{a}$ and $\vec{b}$ ($0°\le \theta \le 180°$)
• $\hat{n}$ is the unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ (right-hand rule)

Component Form (Determinant)

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

From the definition:

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

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

Geometrical Interpretation

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

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Worked Example

Find the angle between $\vec{a}=2\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ using the cross product.

Step 1: Compute $\vec{a}\times \vec{b}$

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

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

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

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

Step 2: Find $|\vec{a}\times \vec{b}|$

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

Step 3: Find $|\vec{a}|$ and $|\vec{b}|$

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

Step 4: Find $\sin \theta$

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

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

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Unit Vectors and Components (Review)

The standard unit vectors in 3D space are:

• $\hat{i}=(1,0,0)$ along the $x$-axis
• $\hat{j}=(0,1,0)$ along the $y$-axis
• $\hat{k}=(0,0,1)$ along the $z$-axis

Any vector $\vec{v}=v_1\hat{i}+v_2\hat{j}+v_3\hat{k}$ has magnitude:

$|\vec{v}|=\sqrt{v_1^2+v_2^2+v_3^2}$

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Properties of Cross Product