3.2 Q-6

Exercise 3.2 — Question 6

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 to each other.

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Key Concepts

Dot Product in Component Form (SLO M-11-B-07):

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

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

Condition for Orthogonality (SLO M-11-B-08):

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

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

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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 perpendicularity, set $\vec{a}\cdot \vec{b}=0$:

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

$=6-2k-4$

$=2-2k$

Setting equal to zero:

$2-2k=0$

$\boxed{k=1}$

Verification: With $k=1$: $\vec{a}\cdot \vec{b}=6-2(1)-4=0$ ✓

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Related Formulas

Angle Between Two Vectors (SLO M-11-B-09):

$\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$

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

Projection of $\vec{a}$ along $\vec{b}$ (SLO M-11-B-10):

$\text{Scalar projection}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$

$\text{Vector projection}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}{|}^2}\vec{b}$

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