Exercise 3.2 — Question 5

Problem Statement

Find the angle between the vectors $\vec{a}=3\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+2\hat{j}-4\hat{k}$. Also find the projection of $\vec{a}$ along $\vec{b}$.

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Key Concepts

Dot Product in Component Form

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

$\vec{a}\cdot \vec{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$

Angle Between Two Vectors

The angle $\theta$ between two vectors is found using:

$\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$

$\theta ={\cos }^{-1}\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)$

Projection of a Vector Along Another

The scalar projection (component) of $\vec{a}$ along $\vec{b}$ is:

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

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Solution

Step 1: Identify the vectors

$\vec{a}=3\hat{i}-\hat{j}+2\hat{k},\vec{b}=2\hat{i}+2\hat{j}-4\hat{k}$

Step 2: Compute the dot product

$\vec{a}\cdot \vec{b}=(3)(2)+(-1)(2)+(2)(-4)=6-2-8=-4$

Step 3: Find the magnitudes

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

$|\vec{b}|=\sqrt{2^2+2^2+(-4{)}^2}=\sqrt{4+4+16}=\sqrt{24}=2\sqrt{6}$

Step 4: Find the angle

$\cos \theta =\frac{-4}{\sqrt{14}\cdot 2\sqrt{6}}=\frac{-4}{2\sqrt{84}}=\frac{-4}{4\sqrt{21}}=\frac{-1}{\sqrt{21}}$

$\theta ={\cos }^{-1}\left(\frac{-1}{\sqrt{21}}\right)$

Step 5: Find the projection of $\vec{a}$ along $\vec{b}$

${\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\frac{-4}{2\sqrt{6}}=\frac{-2}{\sqrt{6}}=\frac{-2\sqrt{6}}{6}=\frac{-\sqrt{6}}{3}$

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Summary of Results