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  • Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 23 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 23 Maths Textbook Solution.

Answers (1)

Answer: the value \lambda will be 1

Given: find the value of \lambda so that the following lines are perpendicular to each other;

\begin{aligned} &\frac{x-5}{5 \lambda+2}=\frac{2-y}{5}=\frac{1-z}{-1}, \frac{x}{1}=\frac{2 y+1}{4 \lambda}=\frac{1-z}{-3} \\ \end{aligned}

Hint: \begin{aligned} &\cos \theta=0 \\ \end{aligned}  because lines are perpendicular.

Solution: \begin{aligned} &\frac{x-5}{5 \lambda+2}=\frac{y-2}{-5}=\frac{z-1}{1} \\ \end{aligned}

Therefore \begin{aligned} &\cos \theta=0 \\ \end{aligned}

                         \begin{aligned} &\quad=(5 \lambda+2)(1)+(-5)(2 \lambda)+(3)(1)=0 \\ \end{aligned}

                          \begin{aligned} &=5 \lambda+2-10 \lambda+3=0 \\ &=5 \lambda=5 \\ &=\lambda=1 \end{aligned}

so the answer will be ‘1’

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