3.2 Q-15

Exercise 3.2 — Question 15

Problem Statement

Find the value of $k$ so that the vectors $\vec{a}=2\hat{i}+k\hat{j}-\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}+4\hat{k}$ are perpendicular (orthogonal) to each other.

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Key Concept: Orthogonality Condition

Two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular if and only if their dot product equals zero:

$\vec{a}\cdot \vec{b}=0$

In component form, if $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, then:

$\vec{a}\cdot \vec{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$

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Solution

Given: $\vec{a}=2\hat{i}+k\hat{j}-\hat{k},\vec{b}=3\hat{i}-2\hat{j}+4\hat{k}$

For $\vec{a}\perp \vec{b}$, we need $\vec{a}\cdot \vec{b}=0$:

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

$=6-2k-4$

$=2-2k$

Setting equal to zero:

$2-2k=0$

$\boxed{k=1}$

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Verification

With $k=1$: $\vec{a}=2\hat{i}+\hat{j}-\hat{k}$

$\vec{a}\cdot \vec{b}=(2)(3)+(1)(-2)+(-1)(4)=6-2-4=0✓$

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Related Concept: Projection of a Vector

The projection of vector $\vec{a}$ along vector $\vec{b}$ is:

${\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$

This gives the scalar component of $\vec{a}$ in the direction of $\vec{b}$.

Example: Find the projection of $\vec{a}=2\hat{i}+3\hat{j}+6\hat{k}$ along $\vec{b}=\hat{i}+2\hat{j}+2\hat{k}$.

$\vec{a}\cdot \vec{b}=(2)(1)+(3)(2)+(6)(2)=2+6+12=20$

$|\vec{b}|=\sqrt{1^2+2^2+2^2}=\sqrt{9}=3$

${\text{proj}}_{\vec{b}}\vec{a}=\frac{20}{3}$

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