3.2 Q-13

Exercise 3.2 — Question 13

Problem Statement

If $\vec{a}=3\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+3\hat{j}-\hat{k}$, find:

1. The angle $\theta$ between $\vec{a}$ and $\vec{b}$
2. 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 $\vec{a}\cdot \vec{b}$

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

Step 2: Find the magnitudes

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

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

Step 3: Find the angle $\theta$

$\cos \theta =\frac{1}{\sqrt{14}\cdot \sqrt{14}}=\frac{1}{14}$

$\theta ={\cos }^{-1}\left(\frac{1}{14}\right)\approx 85.9^{\circ }$

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

${\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\frac{1}{\sqrt{14}}$

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Summary of Results