3.2 Q-4

Exercise 3.2 — Question 4

Problem Statement

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

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Key Concepts

Unit Vectors and Components

Any vector in 3D space can be written in terms of unit vectors $\hat{i}$, $\hat{j}$, $\hat{k}$ along the $x$-, $y$-, and $z$-axes respectively: $\vec{v}=v_1\hat{i}+v_2\hat{j}+v_3\hat{k}$ where $v_1$, $v_2$, $v_3$ are the scalar components.

Magnitude of a Vector $|\vec{v}|=\sqrt{v_1^2+v_2^2+v_3^2}$

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

Angle Between Two Vectors $\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$

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Solution

Step 1: Identify components

$\vec{a}=2\hat{i}-\hat{j}+\hat{k}⟹a_1=2,a_2=-1,a_3=1$ $\vec{b}=-\hat{i}+\hat{j}+0\hat{k}⟹b_1=-1,b_2=1,b_3=0$

Step 2: Compute the dot product

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

Step 3: Compute the magnitudes

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

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

Step 4: Apply the angle formula

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

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

Answer: The angle between $\vec{a}$ and $\vec{b}$ is $\boxed{150°}$.

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