3.4 Q-2

Exercise 3.4 — Question 2

Cross Product: Finding the Angle Between Two Vectors

This question applies the cross product (vector product) to find the angle between two vectors in 3D space.

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Key Formula

The magnitude of the cross product of two vectors $\vec{a}$ and $\vec{b}$ is:

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

where $\theta$ is the angle between the two vectors ($0°\le \theta \le 180°$).

Rearranging to find the angle:

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

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

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Cross Product — Determinant Formula

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

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

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

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Geometrical Interpretation

• $\vec{a}\times \vec{b}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$.
• Its direction is given by the right-hand rule.
• Its magnitude $|\vec{a}\times \vec{b}|$ equals the area of the parallelogram formed by $\vec{a}$ and $\vec{b}$.
• If $\vec{a}\times \vec{b}=\vec{0}$, the vectors are parallel (or one is the zero vector).

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Step-by-Step Method

Step 1: Write both vectors in component form.

Step 2: Compute $\vec{a}\times \vec{b}$ using the determinant formula.

Step 3: Find $|\vec{a}\times \vec{b}|$ using: $|\vec{a}\times \vec{b}|=\sqrt{(a_2{b}_3-a_3{b}_2{)}^2+(a_1{b}_3-a_3{b}_1{)}^2+(a_1{b}_2-a_2{b}_1{)}^2}$

Step 4: Find $|\vec{a}|$ and $|\vec{b}|$ using: $|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2},|\vec{b}|=\sqrt{b_1^2+b_2^2+b_3^2}$

Step 5: Apply $\sin \theta =\frac{|\vec{a}\times \vec{b}|}{|\vec{a}||\vec{b}|}$ and solve for $\theta$.

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Worked Example

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

Cross product: $\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 2& 1& -2\end{array}\right|$ $=\hat{i}(2\cdot (-2)-2\cdot 1)-\hat{j}(1\cdot (-2)-2\cdot 2)+\hat{k}(1\cdot 1-2\cdot 2)$ $=\hat{i}(-4-2)-\hat{j}(-2-4)+\hat{k}(1-4)$ $=-6\hat{i}+6\hat{j}-3\hat{k}$

Magnitudes: $|\vec{a}\times \vec{b}|=\sqrt{36+36+9}=\sqrt{81}=9$ $|\vec{a}|=\sqrt{1+4+4}=3,|\vec{b}|=\sqrt{4+1+4}=3$

Angle: $\sin \theta =\frac{9}{3\times 3}=1⟹\theta =90°$