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  • Provide Solution for RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Very Short Answer Question 9

Provide Solution for RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Very Short Answer Question 9

Answers (1)

Answer:

Required answer is 

Hint:

Use properties of vector

Given:

\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z-2}{1} \text { and } \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{3}

Solution:

We have,

\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z-2}{1} \text { and } \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{3}

The given lines are parallel to the vectors

\begin{aligned} &\vec{b}_{1}=7 \hat{i}-5 \hat{j}+\hat{k} \\ \end{aligned}

\overline{b_{2}}=\hat{i}+2 \hat{j}+3 \hat{k}

Let  \theta  be the angle between the given lines

Now,

\begin{aligned} &\cos \theta=\frac{\vec{b}_{1}\cdot \overline{b_{2}}}{\left|\vec{b}_{1}\right|\left|\overrightarrow{b_{2}}\right|} \\ & \end{aligned}\begin{aligned} &\cos \theta=\frac{\vec{b}_{1}\cdot \overline{b_{2}}}{\left|\vec{b}_{1}\right|\left|\overrightarrow{b_{2}}\right|} \\ & \end{aligned}

=\frac{(7 \hat{i}-5 \hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}+3 \hat{k})}{\sqrt{7^{2}+(-5)^{2}+1^{2}} \cdot \sqrt{1^{2}+2^{2}+3^{2}}}

\begin{aligned} &=\frac{7-10+3}{\sqrt{49+25+1} \cdot \sqrt{1+4+9}} \\ & \end{aligned}

=0 \\

\therefore \theta=\frac{\pi}{2}

Posted by

infoexpert27

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