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

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

Answers (1)

Answer:

Required answer is   \lambda = \frac{-10}{7}

Hint:

Use equation of line in space

Given:

\frac{x-3}{-3}=\frac{y+2}{2 \lambda}=\frac{z+4}{2} \text { And } \frac{x+1}{3 \lambda}=\frac{y-2}{1}=\frac{z+6}{-5}

Solution:

We have,

\frac{x-3}{-3}=\frac{y+2}{2 \lambda}=\frac{z+4}{2} \text { And } \frac{x+1}{3 \lambda}=\frac{y-2}{1}=\frac{z+6}{-5}

The given lines are parallel to vectors

\vec{b}_{1}=-3 \hat{i}+2 \lambda \hat{j}+2 \hat{k} \text { and } \vec{b_{2}}=3 \lambda \hat{i}+\hat{j}-5 \hat{k}

Now, for  \vec{b}_{1} \perp \vec{b_{2}}  we must have,

\begin{aligned} &\vec{b}_{1} \cdot \vec{b_{2}}=0 \\ & \end{aligned}

(-3 \hat{i}+2 \lambda \hat{j}+2 \hat{k}) \cdot(3 \lambda \hat{i}+\hat{j}-5 \hat{k})=0 \\

-7 \lambda-10=0 \\

\lambda=-\frac{10}{7}

 

Posted by

infoexpert27

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