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  • Please solve RD Sharma class 12 chapter 27 Straight Line in Space exercise Multiple choice question 1 maths textbook solution

Please solve RD Sharma class 12 chapter 27 Straight Line in Space exercise Multiple choice question 1 maths textbook solution

Answers (1)

Answer: Correct answer is D.

Hint: Use vector dot product

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

 

Solution: The direction ratios of the given lines are proportional to  2,5,4  and  1,2,-3 . The given lines are parallel to the vectors \Rightarrow \overrightarrow{b_{1}}=2 \hat{\imath}+5 \hat{\jmath}+4 \hat{k} \text { and } \overrightarrow{b_{2}}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k}

Let θ be the angle between the given lines.

Now,

\begin{aligned} &\operatorname{Cos} \theta=\frac{\overrightarrow{b_{1}} \cdot \overrightarrow{b_{2}}}{\left|\overrightarrow{b_{1}}\right|\left|\overrightarrow{b_{2}}\right|} \\\\ &\operatorname{Cos} \theta=\frac{(2 \hat{\imath}+5 \hat{\jmath}+4 \hat{k}) \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})}{\sqrt{2^{2}+5^{2}+4^{2}} \sqrt{1^{2}+2^{2}+(-3)^{2}}} \end{aligned}

\operatorname{Cos} \theta=\frac{2+10-12}{\sqrt{45} \sqrt{14}}

\operatorname{Cos} \theta=

Hence, \theta=\frac{\pi}{2}

\theta=90^{\circ}

So, the correct option is (d) 90°

 

Posted by

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