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  • Explain solution RD Sharma class 12 chapter 27 Straight Line in Space exercise Multiple choice question 4 maths

Explain solution RD Sharma class 12 chapter 27 Straight Line in Space exercise Multiple choice question 4 maths

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Answer: Correct answer is C.

Hint: Use vector dot product.

 

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

 

Solution: The direction ratios of the given lines are proportional to 1 ,1, 2 and --\sqrt{3}-1, \sqrt{3}-1,4

The given lines are parallel to the vectors

\Rightarrow \overrightarrow{b_{1}}=\hat{\imath}+\hat{\jmath}+2 \widehat{k} \text { and } \overrightarrow{b_{2}}=(-\sqrt{3}-1) \hat{\imath}+(\sqrt{3}-1) \hat{\jmath}+4 \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{(\hat{\imath}+\hat{\jmath}+2 \hat{k}) \cdot\{(-\sqrt{3}-1) \hat{\imath}+(\sqrt{3}-1) \hat{\jmath}+4 \hat{k}\}}{\sqrt{1^{2}+1^{2}+2^{2}} \sqrt{(-\sqrt{3}-1)^{2}+(\sqrt{3}-1)^{2}+(4)^{2}}} \end{aligned}

\begin{aligned} &\operatorname{Cos} \theta=\frac{-\sqrt{3}-1+\sqrt{3}-1+8}{\sqrt{3} \sqrt{24}} \\\\ &\operatorname{Cos} \theta=\frac{6}{6 \sqrt{2}} \end{aligned}

\begin{aligned} &\operatorname{Cos} \theta=\frac{1}{\sqrt{2}} \\\\ &\theta=\frac{\pi}{3} \end{aligned}

So, the correct option is  \text { (c) } \frac{\pi}{3}

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