3.2 Q-11

Exercise 3.2 — Question 11

Problem Statement

Find the angle between the vectors $\vec{a}$ and $\vec{b}$, and find the projection of $\vec{a}$ along $\vec{b}$.

---

Key Formulas

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: $|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$

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

---

Worked Solution (General Method)

Let $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$.

Step 1 — Compute the dot product: $\vec{a}\cdot \vec{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$

Step 2 — Compute the magnitudes: $|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2},|\vec{b}|=\sqrt{b_1^2+b_2^2+b_3^2}$

Step 3 — Find the angle: $\theta ={\cos }^{-1}\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)$

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

---

Important Notes

• If $\vec{a}\cdot \vec{b}=0$, the vectors are orthogonal ($\theta =90^{\circ }$).
• The projection is a scalar quantity; it can be negative if the angle is obtuse.
• Unit vectors $\hat{i}$, $\hat{j}$, $\hat{k}$ satisfy: $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1$ and $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0$.