Exercise 3.4 — Question 5

Cross Product and Angle Between Vectors

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

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

Component Form (Determinant)

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

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

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

Finding the Angle Using Cross Product

From the definition:

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

Steps:

1. Compute $\vec{a}\times \vec{b}$ using the determinant formula.
2. Find $|\vec{a}\times \vec{b}|=\sqrt{(a_2{b}_3-a_3{b}_2{)}^2+(a_1{b}_3-a_3{b}_1{)}^2+(a_1{b}_2-a_2{b}_1{)}^2}$.
3. Find $|\vec{a}|$ and $|\vec{b}|$ using $|\vec{v}|=\sqrt{v_1^2+v_2^2+v_3^2}$.
4. Substitute into $\sin \theta =\frac{|\vec{a}\times \vec{b}|}{|\vec{a}||\vec{b}|}$ and solve for $\theta$.

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Worked Example

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

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

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

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

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

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

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

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

Step 3: Find magnitudes

$|\vec{a}|=\sqrt{1+1+1}=\sqrt{3},|\vec{b}|=\sqrt{4+1+1}=\sqrt{6}$

Step 4: Find $\theta$

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

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

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Geometric Interpretation

The magnitude $|\vec{a}\times \vec{b}|$ 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 both vectors.