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  • Explain solution RD Sharma class 12 chapter 27 Straight Line in Space exercise Multiple choice question 16 maths

Explain solution RD Sharma class 12 chapter 27 Straight Line in Space exercise Multiple choice question 16 maths

Answers (1)

Answer: Correct answer is d.

Hint: Use Matrix Product

 

Given:   \frac{x-2}{1}=\frac{y-3}{1}=\frac{4-z}{k} \text { and } \frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{-2}

 

Solution:

\begin{aligned} &\Rightarrow \frac{x-2}{1}=\frac{y-3}{1}=\frac{4-z}{k} \ldots(i) \\ &\text { and } \frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{-2} \ldots(i i) \end{aligned}

Obviously parallel vectors b1 and b2 of line (i) and respectively are:

\begin{aligned} &\Rightarrow \overrightarrow{b_{1}}=-1 \hat{\imath}-2 k \hat{\jmath}+2 \hat{k} \\\\ &\Rightarrow \overrightarrow{b_{2}}=k \hat{\jmath}=2 \hat{\jmath}-2 \hat{k} \\\\ &\text { Lines }(i) \perp(i i) \Rightarrow \overrightarrow{b_{1}} \perp \overrightarrow{b_{2}} \end{aligned}

\begin{aligned} &\Rightarrow \overrightarrow{b_{1}} \cdot \overrightarrow{b_{2}}=0 \\\\ &\Rightarrow-3 k+2 k+10=0 \\\\ &\Rightarrow-5 k+10=0 \end{aligned}

\begin{aligned} &\Rightarrow k=\frac{-10}{-5} \\\\ &\Rightarrow=2(d) \end{aligned}

So, the correct option is  (d)2

Posted by

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