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Vector Algebra question

2025 · 2 Apr · Shift 1 · Q38

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JEE Main
Mathematics
Vector Algebra
2025 · 2 Apr · Shift 1 · Q38

Vector Algebra question

2025 · 2 Apr · Shift 1 · Q38

JEE MainMathematicsVector AlgebraMCQ+4 / −1

If

\overrightarrow{\mathrm{a}}

is a 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

\overrightarrow{\mathrm{a}}

is :

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

View written solutionFree

Correct answer: D

Let

\vec a = x\hat i + y\hat j + z\hat k.

The given vectors are

\vec b_1 = 2\hat i - \hat j + 2\hat k, \qquad \vec b_2 = \hat i + 2\hat j - 2\hat k, \qquad \vec b_3 = \hat k.

The projection of $\vec a$ on a vector $\vec b$ is proportional to

\frac{\vec a\cdot \vec b}{|\vec b|}.

Since the projections on $\vec b_1, \vec b_2,$ and $\hat k$ are equal, we have

\frac{\vec a\cdot \vec b_1}{|\vec b_1|}= rac{\vec a\cdot \vec b_2}{|\vec b_2|}= rac{\vec a\cdot \hat k}{|\hat k|}.

Now,

|\vec b_1|=\sqrt{2^2+(-1)^2+2^2}=3,

|\vec b_2|=\sqrt{1^2+2^2+(-2)^2}=3,

|\hat k|=1.

So,

\frac{2x-y+2z}{3} = \frac{x+2y-2z}{3} = z.

First equate the first two expressions:

2x-y+2z = x+2y-2z.

This gives

x-3y+4z=0. \qquad (1)

Next equate the first expression to $z$ :

\frac{2x-y+2z}{3}=z

2x-y+2z=3z

2x-y-z=0. \qquad (2)

Solve equations (1) and (2):

From (2),

y=2x-z.

Substitute into (1):

x-3(2x-z)+4z=0

x-6x+3z+4z=0

-5x+7z=0

7z=5x

z=\frac{5x}{7}.

Then

y=2x-\frac{5x}{7}=\frac{14x-5x}{7}=\frac{9x}{7}.

Hence,

\vec a \parallel 7\hat i+9\hat j+5\hat k.

Therefore, a unit vector along $\vec a$ is

\frac{1}{\sqrt{7^2+9^2+5^2}}(7\hat i+9\hat j+5\hat k) =\frac{1}{\sqrt{49+81+25}}(7\hat i+9\hat j+5\hat k) =\frac{1}{\sqrt{155}}(7\hat i+9\hat j+5\hat k).

Checking options:

A: incorrect sign pattern
B: incorrect sign pattern
C: incorrect sign pattern
D: matches exactly

Therefore, the correct option is

\boxed{\text{D } \frac{1}{\sqrt{155}}(7\hat i+9\hat j+5\hat k)}.

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