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  • Provide Solution for RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Very Short Answer Question 21

Provide Solution for RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Very Short Answer Question 21

Answers (1)

Answer:

Required angle is  \frac{\pi}{2}

Hint:

Use equation of a line in space.

Given:

 2 x=3 y=-z \text { and } 6 x=-y=-4 z

Solution:

The equation of the given lines can be rewritten as,

\frac{x}{3}=\frac{y}{2}=\frac{z}{-6} \text { and } \frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}

We know that angle between the lines

\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}} \text { and } \frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}    is given by,

\cos \theta=a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}

Let,

\theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}

Let  \theta  be the angle between the given lines

\begin{aligned} &\cos \theta=\frac{3 \times 2+2 \times(-12)+(-6) \times(-3)}{\sqrt{3^{2}+2^{2}+(-6)^{2}} \sqrt{2^{2}+(-12)^{2}+(-3)^{2}}} \\ & \end{aligned}

=\frac{6-24+18}{\sqrt{49} \sqrt{157}}

\begin{aligned} &=0 \\ & \end{aligned}

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

Thus, the angle between the given lines is  \frac{\pi}{2}

 

Posted by

infoexpert27

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