3.2 Q-2

Exercise 3.2 — Question 2

Problem

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

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

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

---

Key Concepts

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

Magnitude of a vector $\vec{v}=v_1\hat{i}+v_2\hat{j}+v_3\hat{k}$: $|\vec{v}|=\sqrt{v_1^2+v_2^2+v_3^2}$

Angle between two vectors: $\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$

Orthogonality condition: Two vectors are perpendicular (orthogonal) if and only if: $\vec{a}\cdot \vec{b}=0$

Scalar projection of $\vec{b}$ along $\vec{a}$: ${\text{proj}}_{\vec{a}}\vec{b}=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}|}$

---

Solution

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

Step 1 — Compute the dot product: $\vec{a}\cdot \vec{b}=(1)(2)+(1)(-1)+(1)(3)=2-1+3=4$

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

Step 3 — Find the angle: $\cos \theta =\frac{4}{\sqrt{3}\cdot \sqrt{14}}=\frac{4}{\sqrt{42}}$ $\theta ={\cos }^{-1}\left(\frac{4}{\sqrt{42}}\right)\approx 51.9°$

---

Part (ii): $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-2\hat{j}+4\hat{k}$

Step 1 — Compute the dot product: $\vec{a}\cdot \vec{b}=(2)(1)+(3)(-2)+(-1)(4)=2-6-4=-8$

Step 2 — Find the magnitudes: $|\vec{a}|=\sqrt{2^2+3^2+(-1{)}^2}=\sqrt{4+9+1}=\sqrt{14}$ $|\vec{b}|=\sqrt{1^2+(-2{)}^2+4^2}=\sqrt{1+4+16}=\sqrt{21}$

Step 3 — Find the angle: $\cos \theta =\frac{-8}{\sqrt{14}\cdot \sqrt{21}}=\frac{-8}{\sqrt{294}}=\frac{-8}{7\sqrt{6}}$ $\theta ={\cos }^{-1}\left(\frac{-8}{7\sqrt{6}}\right)\approx 117.8°$

---

Summary of Results

| Part | $\vec{a}\cdot \vec{b}$ | $|\vec{a}|$ | $|\vec{b}|$ | $\theta$ | |------|------------------------|------------|------------|----------| | (i) | $4$ | $\sqrt{3}$ | $\sqrt{14}$ | $\approx 51.9°$ | | (ii) | $-8$ | $\sqrt{14}$ | $\sqrt{21}$ | $\approx 117.8°$ |