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JEE Mathematics 2025 Question with Solution

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

  • A

    1155(7i^+9j^+5k^)\frac{1}{\sqrt{155}} (7\hat{i} + 9\hat{j} + 5\hat{k})

  • B

    1155(7i^+9j^−5k^)\frac{1}{\sqrt{155}} (7\hat{i} + 9\hat{j} - 5\hat{k})

  • C

    1155(7i^+9j^+5k^)\frac{1}{\sqrt{155}} (7\hat{i} + 9\hat{j} + 5\hat{k})

  • D

    1155(7i^+9j^−5k^)\frac{1}{\sqrt{155}} (7\hat{i} + 9\hat{j} - 5\hat{k})

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: a⃗\vec{a} is a non-zero vector whose projections on b⃗1=2i^−j^+2k^\vec{b}_1 = 2\hat{i} - \hat{j} + 2\hat{k}, b⃗2=i^+2j^−2k^\vec{b}_2 = \hat{i} + 2\hat{j} - 2\hat{k}, and b⃗3=k^\vec{b}_3 = \hat{k} are equal.

Find: A unit vector along a⃗\vec{a}.

The projection of a⃗\vec{a} on a non-zero vector b⃗\vec{b} is

a⃗⋅b⃗∣b⃗∣\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

Let

a⃗=xi^+yj^+zk^\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}

Then

b⃗1=2i^−j^+2k^,b⃗2=i^+2j^−2k^,b⃗3=k^\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}

Their magnitudes are

∣b⃗1∣=22+(−1)2+22=3|\vec{b}_1| = \sqrt{2^2 + (-1)^2 + 2^2} = 3 ∣b⃗2∣=12+22+(−2)2=3|\vec{b}_2| = \sqrt{1^2 + 2^2 + (-2)^2} = 3 ∣b⃗3∣=1|\vec{b}_3| = 1

Since the projections are equal,

a⃗⋅b⃗13=a⃗⋅b⃗23=z\frac{\vec{a} \cdot \vec{b}_1}{3} = \frac{\vec{a} \cdot \vec{b}_2}{3} = z

So,

2x−y+2z3=z\frac{2x - y + 2z}{3} = z

which gives

2x−y=z⋯(I)2x - y = z \qquad \cdots (I)

and

x+2y−2z3=z\frac{x + 2y - 2z}{3} = z

which gives

x+2y=5z⋯(II)x + 2y = 5z \qquad \cdots (II)

Now multiply equation (I)(I) by 22 and add to equation (II)(II):

2(2x−y)+(x+2y)=2z+5z2(2x - y) + (x + 2y) = 2z + 5z 5x=7z5x = 7z x=75zx = \frac{7}{5}z

Substitute into equation (I)(I):

2(75z)−y=z2\left(\frac{7}{5}z\right) - y = z 145z−y=z\frac{14}{5}z - y = z y=95zy = \frac{9}{5}z

Hence,

a⃗=(75z)i^+(95z)j^+zk^=z5(7i^+9j^+5k^)\vec{a} = \left(\frac{7}{5}z\right)\hat{i} + \left(\frac{9}{5}z\right)\hat{j} + z\hat{k} = \frac{z}{5}(7\hat{i} + 9\hat{j} + 5\hat{k})

Therefore, a direction vector along a⃗\vec{a} is 7i^+9j^+5k^7\hat{i} + 9\hat{j} + 5\hat{k}. Its magnitude is

72+92+52=49+81+25=155\sqrt{7^2 + 9^2 + 5^2} = \sqrt{49 + 81 + 25} = \sqrt{155}

So the required unit vector is

1155(7i^+9j^+5k^)\frac{1}{\sqrt{155}}(7\hat{i} + 9\hat{j} + 5\hat{k})

Therefore, the correct option is C.

Common mistakes

  • Using dot products directly without dividing by the magnitudes of the given vectors. Projection on b⃗\vec{b} is a⃗⋅b⃗∣b⃗∣\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}, not just a⃗⋅b⃗\vec{a}\cdot\vec{b}. Always account for the denominator before equating projections.

  • Assuming equal projections imply a⃗⋅b⃗1=a⃗⋅b⃗2=a⃗⋅b⃗3\vec{a}\cdot\vec{b}_1 = \vec{a}\cdot\vec{b}_2 = \vec{a}\cdot\vec{b}_3 without checking that the magnitudes differ. Here ∣b⃗1∣=∣b⃗2∣=3|\vec{b}_1| = |\vec{b}_2| = 3 but ∣b⃗3∣=1|\vec{b}_3| = 1, so the third equation must be handled separately.

  • Finding the direction vector 7i^+9j^+5k^7\hat{i} + 9\hat{j} + 5\hat{k} and stopping there. The question asks for a unit vector, so you must divide by its magnitude 155\sqrt{155}.

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