JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(\mathrm{L}_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}\) and \(\mathrm{L}_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}\) 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 \(\mathrm{L}_1\), then \(|5 \alpha-11 \beta-8 \gamma|\) equals :
- A 20
- B 18
- C 25
- D 16
Answer & Solution
Correct Answer
(C) 25
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { 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{\mathrm{i}}-3…
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