Let L1:x-1/1=y-2/-1=z-1/2 and L2:x+1/-1=y-2/2=z/1 be two lines.

Question

Let \(L_{1}: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}\) and \(\mathrm{L}_{2}: \frac{\mathrm{x}+1}{-1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}}{1}\) be two lines. Let \(L_{3}\) be a line passing through the point \((\alpha, \beta, \gamma)\) and be perpendicular to both \(L_{1}\) and \(L_{2}.\) If \(L_{3}\) intersects \(L_{1},\) then \(|5 \alpha-11 \beta-8 \gamma|\) equals :

(1) 18

(2) 16

(3) 25

(4) 20

Answer 1

Correct option is (3) 25   

DR's of \( L_{3}=\overrightarrow{\mathrm{m}} \times \overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 2 \\ -1 & 2 & 1\end{array}\right|\)

\(=-5 \hat{i}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)

\(L_{3}: \frac{x-\alpha}{-5}=\frac{y-\beta}{-3}=\frac{z-\gamma}{1}=\lambda\)

\(\mathrm{A}(\alpha-5 \lambda, \beta-3 \lambda, \gamma+\lambda)\)

\(\mathrm{L}_{1}: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-1}{2}=\mathrm{k}\)

\(\mathrm{B}(\mathrm{k}+1,-\mathrm{k}+2,2 \mathrm{k}+1)\)

Now

\( \alpha-5 \lambda=\mathrm{k}+1 \Rightarrow \alpha=5 \lambda+\mathrm{k}+1 \)

\(\beta-3 \lambda=-\mathrm{k}+2 \Rightarrow \beta=3 \lambda-\mathrm{k}+2 \)

\(\gamma+\lambda=2 \mathrm{k}+1 \Rightarrow \gamma=-\lambda+2 \mathrm{k}+1 \)

\( |5 \alpha-11 \beta-8 \gamma|=|-25| \)    

= 25