If the components of along and perpendicular to respectively, are and , then is equal to : [2025]

Question

If the components of $\vec{a} = α \hat{i} + β \hat{j} + γ \hat{k}$ along and perpendicular to $\vec{b} = 3 \hat{i} + \hat{j} - \hat{k}$ respectively, are $\frac{16}{11} (3 \hat{i} + \hat{j} - \hat{k})$ and $\frac{1}{11} (- 4 \hat{i} - 5 \hat{j} - 17 \hat{k})$ , then $α^{2} + β^{2} + γ^{2}$ is equal to : [2025]

Smart Achievers · Accepted Answer

(1)

Let ${\vec{a}}_{1}$ = Component of $\vec{a}$ along $\vec{b}$ and ${\vec{a}}_{2}$ = Component of $\vec{a}$ perpendicular to $\vec{b}$

$∴ {\vec{a}}_{1} = \frac{16}{11} (3 \hat{i} + \hat{j} - \hat{k})$ and ${\vec{a}}_{2} = \frac{1}{11} (- 4 \hat{i} - 5 \hat{j} - 17 \hat{k})$

$∵ \vec{a} = {\vec{a}}_{1} + {\vec{a}}_{2}$

$∴ \vec{a} = \frac{16}{11} (3 \hat{i} + \hat{j} - \hat{k}) + \frac{1}{11} (- 4 \hat{i} - 5 \hat{j} - 17 \hat{k})$

$= \frac{44}{11} \hat{i} + \frac{11}{11} \hat{j} - \frac{33}{11} \hat{k} = 4 \hat{i} + \hat{j} - 3 \hat{k}$

On comparing, we get $α = 4, β = 1, γ = - 3$

Hence, $α^{2} + β^{2} + γ^{2} = 16 + 1 + 9 = 26$ .

Solution

1 26

2 23

3 18

4 16

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