Let $\mathrm{L}_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{\mathrm{z}+1}{2}$ and $\mathrm{L}_2:…

Let $\mathrm{L}_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{\mathrm{z}+1}{2}$ and $\mathrm{L}_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$ be two given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is
  1. $\frac{-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}}{\sqrt{99}}$
  2. $\frac{-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{5 \sqrt{3}}$
  3. $\frac{-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{5 \sqrt{3}}$
  4. $\frac{7 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}}{\sqrt{99}}$

Solution

Vector perpendicular to $\mathrm{L}_1$ and $\mathrm{L}_2$ is $\mathrm{L}_1 \times \mathrm{L}_2$ $\begin{aligned} & \therefore \quad \mathrm{L}_1 \times \mathrm{L}_2=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{array}\right|=-\hat{\mathrm{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathrm{k}} \\ & \therefore \quad \text { Required unit vector }=\frac{-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{\sqrt{(-1)^2+(-7)^2+(5)^2}} \\ & =\frac{-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{\sqrt{75}}=\frac{-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{5 \sqrt{3}} \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

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