3.3 Q-12

Exercise 3.3 — Question 12

Problem Statement

Using the cross product, prove the Lagrange Identity:

$|\mathbf{a}\times \mathbf{b}{|}^2=|\mathbf{a}{|}^2|\mathbf{b}{|}^2-(\mathbf{a}\cdot \mathbf{b}{)}^2$

and hence find the sine of the angle between two vectors.

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Key Formulas

Cross product magnitude: $|\mathbf{a}\times \mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin \theta$

Dot product: $\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos \theta$

Angle using cross product: $\sin \theta =\frac{|\mathbf{a}\times \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|}$

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Proof of the Lagrange Identity

Starting from the definitions:

$|\mathbf{a}\times \mathbf{b}{|}^2=|\mathbf{a}{|}^2|\mathbf{b}{|}^2{\sin }^2\theta$

$(\mathbf{a}\cdot \mathbf{b}{)}^2=|\mathbf{a}{|}^2|\mathbf{b}{|}^2{\cos }^2\theta$

Adding:

$|\mathbf{a}\times \mathbf{b}{|}^2+(\mathbf{a}\cdot \mathbf{b}{)}^2=|\mathbf{a}{|}^2|\mathbf{b}{|}^2({\sin }^2\theta +{\cos }^2\theta )=|\mathbf{a}{|}^2|\mathbf{b}{|}^2$

Rearranging:

$\boxed{|\mathbf{a}\times \mathbf{b}{|}^2=|\mathbf{a}{|}^2|\mathbf{b}{|}^2-(\mathbf{a}\cdot \mathbf{b}{)}^2}$

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Geometrical Interpretation

The magnitude of the cross product $|\mathbf{a}\times \mathbf{b}|$ equals the area of the parallelogram formed by vectors $\mathbf{a}$ and $\mathbf{b}$.

• Area of parallelogram $=|\mathbf{a}\times \mathbf{b}|$
• Area of triangle $=\frac{1}{2}|\mathbf{a}\times \mathbf{b}|$

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Worked Example

Find the sine of the angle between $\mathbf{a}=\hat{i}+2\hat{j}+2\hat{k}$ and $\mathbf{b}=3\hat{i}+0\hat{j}+4\hat{k}$.

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

$\mathbf{a}\times \mathbf{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 3& 0& 4\end{array}\right|$

$=\hat{i}(2\cdot 4-2\cdot 0)-\hat{j}(1\cdot 4-2\cdot 3)+\hat{k}(1\cdot 0-2\cdot 3)$

$=\hat{i}(8)-\hat{j}(-2)+\hat{k}(-6)=8\hat{i}+2\hat{j}-6\hat{k}$

Step 2: $|\mathbf{a}\times \mathbf{b}|=\sqrt{64+4+36}=\sqrt{104}=2\sqrt{26}$

Step 3: $|\mathbf{a}|=\sqrt{1+4+4}=3$, $|\mathbf{b}|=\sqrt{9+0+16}=5$

Step 4: $\sin \theta =\frac{2\sqrt{26}}{3\times 5}=\frac{2\sqrt{26}}{15}$

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