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  • Need solution for RD Sharma Maths Class 12 Chapter 27 Straight Line of Space Excercise Very Short Anwer Question 11

Need solution for RD Sharma Maths Class 12 Chapter 27 Straight Line of Space Excercise Very Short Anwer Question 11

Answers (1)

Answer:

Required answer is   \theta = \frac{\pi}{2}

Hint:

Use formula  \cos \theta=\frac{\vec{b}_{1}\cdot \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|}

Given:

2 x=3 y=-z \text { and } 6 x=-y=-4 z

Solution:

We have,

2 x=3 y=-z \text { and } 6 x=-y=-4 z

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}

These lines are parallel to vectors  \vec{b}_{1}=3 \hat{i}+2 \hat{j}-6 \hat{k} \text { and } \overline{b_{2}}=2 \hat{i}-12 \hat{j}-3 \hat{k}

Let \theta  be the angle between these lines.

Now,

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

\begin{aligned} &=\frac{6-24+18}{\sqrt{9+4+36} \cdot \sqrt{4+144+9}}=0 \\ & \end{aligned}

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

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