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JEE Mathematics 2023 Question with Solution

If the points with position vectors αi^+10j^+13k^\alpha \hat{i} + 10\hat{j} + 13\hat{k}, 6i^+11j^+11k^6\hat{i} + 11\hat{j} + 11\hat{k}, and 92i^+βj^−8k^\frac{9}{2}\hat{i} + \beta\hat{j} - 8\hat{k} are collinear, then (19α−6β)2(19\alpha - 6\beta)^2 is equal to:

  • A

    4949

  • B

    3636

  • C

    2525

  • D

    1616

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The position vectors are A=αi^+10j^+13k^A = \alpha \hat{i} + 10\hat{j} + 13\hat{k}, B=6i^+11j^+11k^B = 6\hat{i} + 11\hat{j} + 11\hat{k}, and C=92i^+βj^−8k^C = \frac{9}{2}\hat{i} + \beta\hat{j} - 8\hat{k}.

Find: The value of (19α−6β)2(19\alpha - 6\beta)^2.

For collinear points, the vectors AB⃗\vec{AB} and BC⃗\vec{BC} are parallel, so:

AB⃗×BC⃗=0\vec{AB} \times \vec{BC} = 0

Compute the vectors:

AB⃗=(6−α)i^+(11−10)j^+(11−13)k^=(6−α)i^+j^−2k^\vec{AB} = (6 - \alpha)\hat{i} + (11 - 10)\hat{j} + (11 - 13)\hat{k} = (6 - \alpha)\hat{i} + \hat{j} - 2\hat{k} BC⃗=(92−6)i^+(β−11)j^+(−8−11)k^=−32i^+(β−11)j^−19k^\vec{BC} = \left(\frac{9}{2} - 6\right)\hat{i} + (\beta - 11)\hat{j} + (-8 - 11)\hat{k} = -\frac{3}{2}\hat{i} + (\beta - 11)\hat{j} - 19\hat{k}

Now use the determinant form of the cross product:

AB⃗×BC⃗=∣i^j^k^6−α1−2−32β−11−19∣\vec{AB} \times \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 - \alpha & 1 & -2 \\ -\frac{3}{2} & \beta - 11 & -19 \end{vmatrix}

Setting the components equal to zero and solving gives:

α=11719,β=412\alpha = \frac{117}{19}, \quad \beta = \frac{41}{2}

Substitute these values into 19α−6β19\alpha - 6\beta:

19α−6β=19⋅11719−6⋅412=117−123=−619\alpha - 6\beta = 19 \cdot \frac{117}{19} - 6 \cdot \frac{41}{2} = 117 - 123 = -6

Therefore,

(19α−6β)2=(−6)2=36(19\alpha - 6\beta)^2 = (-6)^2 = 36

So, the correct option is B.

Using component ratios for parallel vectors

Given: The three points are collinear.

Find: (19α−6β)2(19\alpha - 6\beta)^2.

If points are collinear, then AB⃗\vec{AB} and BC⃗\vec{BC} are parallel. Hence their corresponding components are proportional.

From the solution data:

AB⃗=(6−α,1,−2),BC⃗=(−32,β−11,−19)\vec{AB} = (6 - \alpha, 1, -2), \quad \vec{BC} = \left(-\frac{3}{2}, \beta - 11, -19\right)

Using the solved values obtained from the parallel-condition equations:

α=11719,β=412\alpha = \frac{117}{19}, \quad \beta = \frac{41}{2}

Now evaluate the required expression directly.

19α−6β=19⋅11719−6⋅412=117−123=−619\alpha - 6\beta = 19 \cdot \frac{117}{19} - 6 \cdot \frac{41}{2} = 117 - 123 = -6

Therefore,

(19α−6β)2=36(19\alpha - 6\beta)^2 = 36

Hence, the answer is 3636.

Common mistakes

  • Using the condition AB⃗=BC⃗\vec{AB} = \vec{BC} instead of checking that they are parallel is incorrect. Collinear points only require the direction vectors to be parallel, not equal. Use AB⃗×BC⃗=0\vec{AB} \times \vec{BC} = 0 or component proportionality instead.

  • Making sign errors while forming BC⃗\vec{BC} is common, especially in the k^\hat{k}-component. Since −8−11=−19-8 - 11 = -19, writing 1919 changes the entire result. Carefully subtract coordinates component-wise.

  • Confusing the option numbering with option labels can lead to the wrong marked answer. Here, source option (2)(2) corresponds to label B, not C. Always map 1→A1 \to A, 2→B2 \to B, 3→C3 \to C, 4→D4 \to D unless the page explicitly labels otherwise.

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