JEE 2025 — Mathematics PYQ

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Question

JEE 2025 — Mathematics PYQ

JEE | Mathematics | 2025

If $a$ is a non-zero vector such that its projections on the vectors $2 \hat{i} - \hat{j} + 2 \hat{k}$ , $\hat{i} + 2 \hat{j} - 2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $a$ is:

Choose the correct answer:

Explanation

Let $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$

$a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 1$

Let $b = 2 \hat{i} - j + 2 k, c = \hat{i} + 2 j - 2 k$

$d = \hat{k}$

Given projections are equal therefore,

$\frac{a \cdot b}{∣ b ∣} = \frac{a \cdot c}{∣ c ∣} = \frac{a \cdot d}{∣ d ∣}$

$\frac{2 a _{1} - a _{2} + 2 a _{3}}{3} = \frac{a _{1} + 2 a _{2} - 2 a _{3}}{3} = a_{3}$

∴ $2 a_{1} - a_{2} + 2 a_{3} = a_{1} + 2 a_{2} - 2 a_{3}$ …(i)

$a_{1} - 3 a_{2} + 4 a_{3} = 0$ …(ii)

$2 a_{1} - a_{2} + 2 a_{3} = 3 a_{3}$ …(iii)

From Eq. (ii), we get

$a_{2} + a_{3} = 2 t$ {Let, $a_{1} = t$ }

From Eq. (i), we get

$3 a_{2} - 4 a_{3} = t$

∴ $3 a_{2} - 4 (2 t - a_{2}) = t$

∴ $7 a_{2} + 4 t - 8 t = t$

∴ $7 a_{2} = 9 t$

∴ $a_{2} = \frac{9 t}{7}$

$a_{2} + a_{3} = 2 t$

$\frac{9 t}{7} + a_{3} = 2 t$

$∴ a_{3} = \frac{5 t}{7}$

Hence, $a_{1} : a_{2} : a_{3} = t : \frac{9 t}{7} : \frac{5 t}{7}$

$= 7 : 9 : 5$

∴ Unit vector along $a = \frac{1}{155} (7 \hat{i} + 9 \hat{j} + 5 \hat{k})$

Choose the correct answer:

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