3.3 Q-10

Exercise 3.3 — Question 10

Problem Statement

Find the sine of the angle between the vectors $\vec{a}=\hat{i}+2\hat{j}-2\hat{k}$ and $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$. Also find the area of the parallelogram and triangle formed by these vectors.

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Key Formulas

The cross product of two vectors gives:

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

Therefore, the sine of the angle between two vectors is:

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

The Lagrange Identity relates the dot and cross products:

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

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Solution

Step 1: Find the magnitudes

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

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

Step 2: Compute the cross product $\vec{a}\times \vec{b}$

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

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

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

$=-2\hat{i}-5\hat{j}-6\hat{k}$

Step 3: Find $|\vec{a}\times \vec{b}|$

$|\vec{a}\times \vec{b}|=\sqrt{(-2{)}^2+(-5{)}^2+(-6{)}^2}=\sqrt{4+25+36}=\sqrt{65}$

Step 4: Find $\sin \theta$

$\sin \theta =\frac{|\vec{a}\times \vec{b}|}{|\vec{a}||\vec{b}|}=\frac{\sqrt{65}}{3\times 3}=\frac{\sqrt{65}}{9}$

Step 5: Area of the parallelogram

$\text{Area of parallelogram}=|\vec{a}\times \vec{b}|=\sqrt{65}\text{ sq. units}$

Step 6: Area of the triangle

$\text{Area of triangle}=\frac{1}{2}|\vec{a}\times \vec{b}|=\frac{\sqrt{65}}{2}\text{ sq. units}$

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Verification using Lagrange Identity

First compute $\vec{a}\cdot \vec{b}$:

$\vec{a}\cdot \vec{b}=(1)(2)+(2)(-2)+(-2)(1)=2-4-2=-4$

Using the Lagrange Identity:

$|\vec{a}\times \vec{b}{|}^2=|\vec{a}{|}^2|\vec{b}{|}^2-(\vec{a}\cdot \vec{b}{)}^2=9\times 9-(-4{)}^2=81-16=65$

$|\vec{a}\times \vec{b}|=\sqrt{65}✓$

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Summary of Results