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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise Multiple choice question 14

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise Multiple choice question 14

Answers (1)

Answer: Correct answer is D.

Hint: use vector Dot product

 

Given: \frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1} \text { and } \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}

 

Solution:

\begin{aligned} &\vec{a}=3 \hat{\imath}+3 \hat{\jmath}+3 \hat{k} \quad, \vec{b}=3 \hat{\imath}-7 \hat{\jmath}+6 \hat{k} \quad, \vec{p}=3 \hat{\imath}-\hat{\jmath}+\hat{k} \\\\ &\vec{q}=-3 \hat{\imath}-2 \hat{\jmath}+4 \hat{k} \end{aligned}

\vec{p} \times \vec{q}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{array}\right|=-6 \hat{\imath}+15 \hat{\jmath}+3 \hat{k}

\begin{aligned} &\text { S. } D=\left|\frac{\mid \overrightarrow{(a)}-(\overrightarrow{b)}(\vec{p} x \vec{q})}{|\vec{p} x \vec{q}|}\right| \\\\ &=\left|\frac{(6 \hat{\imath}+15 \hat{\jmath}+3 \hat{k})}{\sqrt{36+225+9}}\right| \end{aligned}

\begin{aligned} &=\left|\frac{-36-225-9}{\sqrt{36+225+9}}\right| \\\\ &=\left|\frac{-270}{\sqrt{270}}\right| \\\\ &=\sqrt{270} \Rightarrow 3 \sqrt{30} \end{aligned}

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