3.3 Q-13

Exercise 3.3 — Question 13

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 (Vector) Product

The cross product of two vectors $\vec{a}$ and $\vec{b}$ in 3D space 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).

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:

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

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

Magnitude of the Cross Product

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

Finding the Angle Using Cross Product

Rearranging the definition:

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

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

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Worked Example

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

Step 1: Compute the cross product

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

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

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

$=-2\hat{i}+3\hat{j}-\hat{k}$

Step 2: Find the magnitude of the cross product

$|\vec{a}\times \vec{b}|=\sqrt{(-2{)}^2+3^2+(-1{)}^2}=\sqrt{4+9+1}=\sqrt{14}$

Step 3: Find the magnitudes of $\vec{a}$ and $\vec{b}$

$|\vec{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$

$|\vec{b}|=\sqrt{2^2+1^2+(-1{)}^2}=\sqrt{4+1+1}=\sqrt{6}$

Step 4: Find the angle

$\sin \theta =\frac{\sqrt{14}}{\sqrt{3}\cdot \sqrt{6}}=\frac{\sqrt{14}}{\sqrt{18}}=\frac{\sqrt{14}}{3\sqrt{2}}=\sqrt{\frac{7}{9}}=\frac{\sqrt{7}}{3}$

$\theta ={\sin }^{-1}\left(\frac{\sqrt{7}}{3}\right)$

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Important Properties of Cross Product