3.2 Q-12

Exercise 3.2 — Question 12

Problem Statement

Find the angle between the vectors $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}$, and 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: Identify components

$\vec{a}=2\hat{i}-\hat{j}+\hat{k},\vec{b}=-\hat{i}+\hat{j}+0\hat{k}$

Step 2: Compute the dot product

$\vec{a}\cdot \vec{b}=(2)(-1)+(-1)(1)+(1)(0)=-2-1+0=-3$

Step 3: Find the magnitudes

$|\vec{a}|=\sqrt{(2{)}^2+(-1{)}^2+(1{)}^2}=\sqrt{4+1+1}=\sqrt{6}$

$|\vec{b}|=\sqrt{(-1{)}^2+(1{)}^2+(0{)}^2}=\sqrt{1+1}=\sqrt{2}$

Step 4: Find the angle

$\cos \theta =\frac{-3}{\sqrt{6}\cdot \sqrt{2}}=\frac{-3}{\sqrt{12}}=\frac{-3}{2\sqrt{3}}=\frac{-\sqrt{3}}{2}$

$\theta ={\cos }^{-1}\left(\frac{-\sqrt{3}}{2}\right)=150°$

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

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

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Summary