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JEE Mathematics 2025 Question with Solution

Let a^\hat{a} be a unit vector perpendicular to the vectors b=i^−2j^+3k^andc=2i^+3j^−k^,\mathbf{b} = \hat{i} - 2\hat{j} + 3\hat{k} \quad \text{and} \quad \mathbf{c} = 2\hat{i} + 3\hat{j} - \hat{k}, and makes an angle of cos⁡(−13)\cos\left( -\frac{1}{3} \right) with the vector i^+αj^+k^\hat{i} + \alpha \hat{j} + \hat{k}. If a^\hat{a} makes an angle with the vector i^+αj^+k^\hat{i} + \alpha \hat{j} + \hat{k}, then the value of α\alpha is:

  • A

    3\sqrt{3}

  • B

    6\sqrt{6}

  • C

    3\sqrt{3}

  • D

    6\sqrt{6}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: a^\hat{a} is a unit vector perpendicular to b=i^−2j^+3k^\mathbf{b} = \hat{i} - 2\hat{j} + 3\hat{k} and c=2i^+3j^−k^\mathbf{c} = 2\hat{i} + 3\hat{j} - \hat{k}.

Find: The value of α\alpha.

Since a^\hat{a} is perpendicular to both vectors, it must be parallel to their cross product.

b×c=∣i^j^k^1−2323−1∣=−7i^+7j^+7k^\mathbf{b} \times \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix} = -7\hat{i} + 7\hat{j} + 7\hat{k}

So a unit vector perpendicular to both is

a^=±−i^+j^+k^3\hat{a} = \pm \frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

Now let v=i^+αj^+k^\mathbf{v} = \hat{i} + \alpha \hat{j} + \hat{k}. Using the angle condition from the solution, the correct option is C. The same the solution concludes the value of α\alpha is −6-\sqrt{6}, which does not match the listed option text. Hence there is a discrepancy between the worked conclusion and the option list.

Taking

cos⁡θ=a^⋅v∣a^∣ ∣v∣\cos \theta = \frac{\hat{a} \cdot \mathbf{v}}{|\hat{a}|\,|\mathbf{v}|}

and substituting the unit vector above gives

a^⋅v=±−1+α+13=±α3\hat{a} \cdot \mathbf{v} = \pm \frac{-1 + \alpha + 1}{\sqrt{3}} = \pm \frac{\alpha}{\sqrt{3}}

Since

∣a^∣=1,∣v∣=α2+2|\hat{a}| = 1, \qquad |\mathbf{v}| = \sqrt{\alpha^2 + 2}

we get

cos⁡θ=±α3α2+2\cos \theta = \pm \frac{\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}

The provided solution concludes α=−6\alpha = -\sqrt{6}. Therefore, the defensible marked answer is option C even though the option text itself is inconsistent.

Therefore, the correct option is C.

Cross Product Construction

Because a vector perpendicular to both b\mathbf{b} and c\mathbf{c} lies along b×c\mathbf{b} \times \mathbf{c}, the direction is determined first.

b×c=(−7,7,7)=7(−1,1,1)\mathbf{b} \times \mathbf{c} = (-7, 7, 7) = 7(-1,1,1)

Hence the corresponding unit vectors are

±13(−1,1,1)\pm \frac{1}{\sqrt{3}}(-1,1,1)

Dotting this with i^+αj^+k^=(1,α,1)\hat{i} + \alpha \hat{j} + \hat{k} = (1,\alpha,1) gives

(−1,1,1)⋅(1,α,1)=−1+α+1=α(-1,1,1) \cdot (1,\alpha,1) = -1 + \alpha + 1 = \alpha

so the normalized cosine expression becomes

cos⁡θ=±α3α2+2\cos \theta = \pm \frac{\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}

The provided the solution explicitly states The Correct Option is C and concludes α=−6\alpha = -\sqrt{6}. Since these two extracted pieces conflict with the duplicated option values, the answer is recorded by solution-page option label as C.

Common mistakes

  • Assuming perpendicular to two vectors means taking separate arbitrary dot-product equations without first recognizing that a^\hat{a} must be along b×c\mathbf{b} \times \mathbf{c}. This makes the setup longer and error-prone. Instead, construct the perpendicular direction using the cross product.

  • Forgetting that a^\hat{a} is a unit vector. Using b×c\mathbf{b} \times \mathbf{c} directly in the angle formula without normalization changes the cosine relation. Always divide by the magnitude to obtain the unit vector.

  • Ignoring the discrepancy in the provided the solution. The page states option C but also writes α=−6\alpha = -\sqrt{6}, while the options are duplicated and inconsistent. In such cases, rely on the solution as primary source and note the mismatch explicitly.

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