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JEE Mains · Maths · STD 12 - 11. three dimension geometry

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

  1. A \(36\)
  2. B \(16\)
  3. C \(25\)
  4. D \(49\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(36\)

Step-by-step Solution

Detailed explanation

\((\alpha, 10,13) ;(6,11,11),\left(\frac{9}{2}, \beta,-8\right)\) \(\frac{\alpha-6}{3 / 2}=\frac{-1}{11-\beta}=\frac{2}{19} -19=22-2 \beta\) \(\alpha-6=\frac{3}{19} 2 \beta=41\) \(\alpha=6+\frac{3}{19}=\frac{117}{19}\) \(\therefore(19 \alpha-6 \beta)^2=(117-123)^2=36\)
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