JEE Main 2025 Jan 28 Shift 2, Mathematics Q3: Applications of Vectors

JEE Main2025Jan 28, Shift 2Mathematics

If the components of $a = α \hat{i} + β \hat{j} + γ \hat{k}$ along and perpendicular to $b = 3 \hat{i} + \hat{j} - \hat{k}$ 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 :

A
$23$
B
$18$
C
$16$
D
$26$

Solution

Any vector is the sum of its parallel and perpendicular components, so

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

Adding component-wise: $\hat{i}$ -part $= \frac{48 - 4}{11} = 4$ ; $\hat{j}$ -part $= \frac{16 - 5}{11} = 1$ ; $\hat{k}$ -part $= \frac{- 16 - 17}{11} = - 3$ .

So $(α, β, γ) = (4, 1, - 3)$ and $α^{2} + β^{2} + γ^{2} = 16 + 1 + 9 = 26$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q3: Applications of Vectors

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