If a is nonzero vector such that its projections on the vectors 2 i - j +2 k , i +2 j -2 k and k are equal, then a unit vector along a is:

Correct answer: C. 1 155 (7 i +9 j +5 k ) Let a =a_1 i +a_2 j +a_3 k a_1^2+a_2^2+a_3^2=1 Let b =2 i - j +2 k , c = i -2 j -2 k d = k a · b |b| = a · c |c| = a · d |d| 2 a_1-a_2+2 a_3 3 = a_1+2 a_2-2 a_3 3 =a_3 By solving a_1= 7 155 , a_2= 9 155 , a_3= 5 155

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Vector Algebra

If a is nonzero vector such that its projections on the vectors 2 i...

\vec{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

\vec{a}

is:

Options

A $\frac{1}{\sqrt{155}}(-7 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})$
B $\frac{1}{\sqrt{155}}(-7 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$
C $\frac{1}{\sqrt{155}}(7 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})$
D $\frac{1}{\sqrt{155}}(7 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$

Correct answer

C. $\frac{1}{\sqrt{155}}(7 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})$

Step-by-step solution

Let

\overline{\mathrm{a}}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{a}_2 \hat{\mathrm{j}}+\mathrm{a}_3 \hat{\mathrm{k}}

\mathrm{a}_1^2+\mathrm{a}_2^2+\mathrm{a}_3^2=1

Let

\overline{\mathrm{b}}=2 \overrightarrow{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{c}}=\overrightarrow{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}

\overline{\mathrm{d}}=\hat{\mathrm{k}}

\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\mathrm{b}|}=\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}}{|\mathrm{c}|}=\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{d}}}{|\mathrm{d}|}

\frac{2 \mathrm{a}_1-\mathrm{a}_2+2 \mathrm{a}_3}{3}=\frac{\mathrm{a}_1+2 \mathrm{a}_2-2 \mathrm{a}_3}{3}=\mathrm{a}_3

By solving

\mathrm{a}_1=\frac{7}{\sqrt{155}}, \mathrm{a}_2=\frac{9}{\sqrt{155}}, \mathrm{a}_3=\frac{5}{\sqrt{155}}

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