3.2 Q-7

Exercise 3.2 — Question 7

This question applies the dot product to find angles between vectors and projections of one vector along another.

---

Key Formulas

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

Angle Between Two Vectors: $\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|},\theta ={\cos }^{-1}\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)$

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

Orthogonality Condition: $\vec{a}\perp \vec{b}⟺\vec{a}\cdot \vec{b}=0$

---

Worked Solution

Let the vectors given in Q7 be: $\vec{a}=2\hat{i}+3\hat{j}-\hat{k},\vec{b}=\hat{i}-2\hat{j}+4\hat{k}$

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: $\cos \theta =\frac{-8}{\sqrt{14}\cdot \sqrt{21}}=\frac{-8}{\sqrt{294}}$ $\theta ={\cos }^{-1}\left(\frac{-8}{\sqrt{294}}\right)$

Step 4 — Find the projection of $\vec{a}$ along $\vec{b}$: ${\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\frac{-8}{\sqrt{21}}$

---

Orthogonality Check

Two vectors are perpendicular if and only if their dot product equals zero.

Example: For which value of $k$ are $\vec{u}=2\hat{i}+k\hat{j}-\hat{k}$ and $\vec{v}=3\hat{i}-2\hat{j}+4\hat{k}$ perpendicular?

$\vec{u}\cdot \vec{v}=(2)(3)+k(-2)+(-1)(4)=6-2k-4=2-2k=0$ $\therefore k=1$

---