3.3 Q-4

Exercise 3.3 — Question 4

Topic: Cross (Vector) Product

This question involves applying the cross product of two vectors to find a vector perpendicular to two given vectors, or to find the angle between them.

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Key Formula: Cross Product

For two vectors $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, the cross product is:

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

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

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Magnitude of Cross Product

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

where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

Geometrical Interpretation: $|\vec{a}\times \vec{b}|$ equals the area of the parallelogram formed by $\vec{a}$ and $\vec{b}$.

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Finding the Angle Using Cross Product

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

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

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Properties of Cross Product

• $\vec{a}\times \vec{b}=-(\vec{b}\times \vec{a})$ (Anti-commutative)
• $\vec{a}\times \vec{a}=\vec{0}$ (Cross product of a vector with itself is zero)
• $\hat{i}\times \hat{j}=\hat{k},\hat{j}\times \hat{k}=\hat{i},\hat{k}\times \hat{i}=\hat{j}$
• If $\vec{a}\times \vec{b}=\vec{0}$ and $\vec{a},\vec{b}\ne \vec{0}$, then $\vec{a}$ and $\vec{b}$ are parallel.

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Worked Example

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

Step 1: Compute the cross product

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

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

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

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

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

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

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