3.2 Q-1

Exercise 3.2 — Question 1

This exercise covers the dot (scalar) product of vectors in space, including its component form, geometric interpretation, orthogonality, angle between vectors, and projections.

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Key Concepts Reviewed

Dot Product in Component Form

For $\vec{u}=u_1\hat{i}+u_2\hat{j}+u_3\hat{k}$ and $\vec{v}=v_1\hat{i}+v_2\hat{j}+v_3\hat{k}$:

$\vec{u}\cdot \vec{v}=u_1{v}_1+u_2{v}_2+u_3{v}_3$

Geometric Interpretation

$\vec{u}\cdot \vec{v}=|\vec{u}||\vec{v}|\cos \theta$

where $\theta$ is the angle between the two vectors ($0^{\circ }\le \theta \le 180^{\circ }$). The dot product measures how much one vector "projects" onto another, scaled by the magnitudes.

Angle Between Two Vectors

$\theta ={\cos }^{-1}\left(\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}\right)$

Orthogonality Condition

Two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular if and only if: $\vec{a}\cdot \vec{b}=0⟺a_1{b}_1+a_2{b}_2+a_3{b}_3=0$

Projection of a Vector

The scalar projection of $\vec{b}$ along $\vec{a}$: ${\text{comp}}_{\vec{a}}\vec{b}=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}|}$

The vector projection of $\vec{b}$ onto $\vec{a}$: ${\text{proj}}_{\vec{a}}\vec{b}=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}{|}^2}\vec{a}$

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Worked Examples

Example 1: Find $\vec{a}\cdot \vec{b}$ if $\vec{a}=3\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+4\hat{j}-2\hat{k}$.

$\vec{a}\cdot \vec{b}=(3)(1)+(-2)(4)+(1)(-2)=3-8-2=-7$

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

$\vec{a}\cdot \vec{b}=0+1+0=1,|\vec{a}|=\sqrt{2},|\vec{b}|=\sqrt{2}$ $\cos \theta =\frac{1}{\sqrt{2}\cdot \sqrt{2}}=\frac{1}{2}⟹\theta =60^{\circ }$

Example 3: Determine whether $\vec{a}=2\hat{i}-3\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}+\hat{k}$ are orthogonal.

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

Since $\vec{a}\cdot \vec{b}=0$, the vectors are orthogonal (perpendicular).

Example 4: Find the projection of $\vec{b}=2\hat{i}+3\hat{j}-\hat{k}$ along $\vec{a}=\hat{i}-2\hat{j}+2\hat{k}$.

$\vec{a}\cdot \vec{b}=2-6-2=-6,|\vec{a}|=\sqrt{1+4+4}=3$ ${\text{comp}}_{\vec{a}}\vec{b}=\frac{-6}{3}=-2$

The negative sign indicates $\vec{b}$ has a component in the direction opposite to $\vec{a}$.

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