3.2 Q-10

Exercise 3.2 — Question 10

Problem

Find the angle between the vectors $\vec{a}=3\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+2\hat{j}-\hat{k}$. Also find the projection of $\vec{a}$ along $\vec{b}$.

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Key Formulas

Angle between two vectors: $\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$

Projection of $\vec{a}$ along $\vec{b}$: ${\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$

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Solution

Step 1: Compute the dot product $\vec{a}\cdot \vec{b}$

$\vec{a}\cdot \vec{b}=(3)(2)+(-1)(2)+(2)(-1)=6-2-2=2$

Step 2: Find the magnitudes

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

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

Step 3: Find the angle $\theta$

$\cos \theta =\frac{2}{\sqrt{14}\cdot 3}=\frac{2}{3\sqrt{14}}$

$\theta ={\cos }^{-1}\left(\frac{2}{3\sqrt{14}}\right)$

Step 4: Find the projection of $\vec{a}$ along $\vec{b}$

${\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\frac{2}{3}$

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Summary of Results