If the points with vectors $\alpha \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}$, $6 \hat{i}+11 \hat{j}+11 \hat{k}$, $\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}$ are collinear, then $(19 \alpha-6 \beta)^{2}$ is equal to

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Vector Algebra

If the points with vectors

α \hat{i} + 10 \hat{j} + 13 \hat{k}

6 \hat{i} + 11 \hat{j} + 11 \hat{k}

\frac{9}{2} \hat{i} + β \hat{j} - 8 \hat{k}

are collinear, then

(19 α - 6 β)^{2}

is equal to

Answer: (A)

Let

A = (α, 10, 13)

B = (6, 11, 11)

C = (\frac{9}{2}, β, - 8)

. For collinearity,

A B ∥ B C

. We have

A B = (6 - α, 1, - 2)

and

B C = (\frac{9}{2} - 6, β - 11, - 19) = (- \frac{3}{2}, β - 11, - 19)

. For parallel vectors:

\frac{6 - α}{- \frac{3}{2}} = \frac{1}{β - 11} = \frac{- 2}{- 19} = \frac{2}{19}

. From the first ratio:

6 - α = - \frac{3}{2} \cdot \frac{2}{19} = - \frac{3}{19}

, so

α = 6 + \frac{3}{19} = \frac{117}{19}

. From the second ratio:

β - 11 = \frac{19}{2}

, so

β = 11 + \frac{19}{2} = \frac{41}{2}

. Therefore

19 α = 117

and

6 β = 123

, giving

(19 α - 6 β)^{2} = (117 - 123)^{2} = (- 6)^{2} = 36

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option