3.2 Q-3

Exercise 3.2 — Question 3

Problem

Find the dot product of the following pairs of vectors, and determine the angle between them:

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

(ii) $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-\hat{j}-\hat{k}$

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Key Formulas

Dot Product in Component Form: $\vec{a}\cdot \vec{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$

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

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

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Solution

Part (i): $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-2\hat{j}+3\hat{k}$

Step 1 — Compute the dot product: $\vec{a}\cdot \vec{b}=(2)(1)+(3)(-2)+(-1)(3)=2-6-3=-7$

Step 2 — Compute the magnitudes: $|\vec{a}|=\sqrt{2^2+3^2+(-1{)}^2}=\sqrt{4+9+1}=\sqrt{14}$ $|\vec{b}|=\sqrt{1^2+(-2{)}^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$

Step 3 — Find the angle: $\cos \theta =\frac{-7}{\sqrt{14}\cdot \sqrt{14}}=\frac{-7}{14}=-\frac{1}{2}$ $\theta ={\cos }^{-1}\left(-\frac{1}{2}\right)=120°$

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Part (ii): $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-\hat{j}-\hat{k}$

Step 1 — Compute the dot product: $\vec{a}\cdot \vec{b}=(1)(1)+(1)(-1)+(1)(-1)=1-1-1=-1$

Step 2 — Compute the magnitudes: $|\vec{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$ $|\vec{b}|=\sqrt{1^2+(-1{)}^2+(-1{)}^2}=\sqrt{3}$

Step 3 — Find the angle: $\cos \theta =\frac{-1}{\sqrt{3}\cdot \sqrt{3}}=\frac{-1}{3}$ $\theta ={\cos }^{-1}\left(-\frac{1}{3}\right)\approx 109.47°$

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Summary of Concepts Used