Vector Algebra — JEE Main · 2025 · 02 Apr 2025 · Shift 1 Question with Solution

a

is nonzero 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:

Answer: (C)

Solution

Let

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

be a unit vector, so

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

.

Let

b = 2 \hat{i} - \hat{j} + 2 \hat{k}

c = \hat{i} + 2 \hat{j} - 2 \hat{k}

d = \hat{k}

.

Projections are equal:

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

Since

∣ b ∣ = ∣ c ∣ = 3

and

∣ d ∣ = 1

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

From the first two:

2 a_{1} - a_{2} + 2 a_{3} = a_{1} + 2 a_{2} - 2 a_{3} \Rightarrow a_{1} = 3 a_{2} - 4 a_{3}

From the first and third:

2 a_{1} - a_{2} + 2 a_{3} = 3 a_{3} \Rightarrow 2 a_{1} - a_{2} = a_{3}

Solving with

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

gives

a_{1} = \frac{7}{155}

a_{2} = \frac{9}{155}

a_{3} = \frac{5}{155}

.

Unit vector:

\frac{1}{155} (7 \hat{i} + 9 \hat{j} + 5 \hat{k})

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option