JEE Main 2025 Apr 4 Shift 1, Mathematics Q14: Scalar (or Dot) Product of Two Vectors

JEE Main2025Apr 4, Shift 1Mathematics

Consider two vectors $u = 3 \hat{i} - \hat{j}$ and $v = 2 \hat{i} + \hat{j} - λ \hat{k}$ , $λ > 0$ . The angle between them is given by $cos^{- 1} (\frac{5}{2 7})$ . Let $v = v_{1} + v_{2}$ , where $v_{1}$ is parallel to $u$ and $v_{2}$ is perpendicular to $u$ . Then the value $∣ v_{1} ∣^{2} + ∣ v_{2} ∣^{2}$ is equal to

A
$\frac{23}{2}$
B
14
C
$\frac{25}{2}$
D
10

Solution

$u \cdot v = 6 - 1 = 5$ , $∣ u ∣ = 10$ , $∣ v ∣ = 5 + λ^{2}$ .

Using $cos θ = \frac{u \cdot v}{∣ u ∣∣ v ∣} = \frac{5}{2 7}$ : $\frac{5}{10 5 + λ ^{2}} = \frac{5}{2 7}$ .

Squaring: $\frac{25}{10 ( 5 + λ ^{2} )} = \frac{5}{28}$ , giving $5 + λ^{2} = 14$ , so $λ^{2} = 9$ and (since $λ > 0$ ) $λ = 3$ . Thus $∣ v ∣^{2} = 14$ .

By construction $v_{1} ⊥ v_{2}$ , so by the Pythagorean identity for orthogonal decomposition, $∣ v ∣^{2} = ∣ v_{1} ∣^{2} + ∣ v_{2} ∣^{2} = 14$ .

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JEE Main 2025 Apr 4 Shift 1, Mathematics Q14: Scalar (or Dot) Product of Two Vectors

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