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  • Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 21 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 21 Maths Textbook Solution.

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Answer : the value of the \lambda will be \frac{10}{7}

Given: if the lines \frac{x-1}{-3}=\frac{y-2}{2 \lambda}=\frac{z-3}{2} \text { and } \frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{z-6}{5} are perpendicular find the value of \lambda

Hint: Because lines are perpendicular

                So \vec{a}.\vec{b}=0

Solution:   \frac{x-1}{-3}=\frac{y-2}{2 \lambda}=\frac{z-3}{2} \text { and } \frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{z-6}{5} are perpendicular that means they =’0’

                So, \cos \theta =0

                \begin{aligned} &=(3 \lambda)(-3)+(2 \lambda)(1)+(2 \times 5)=0 \\\\ &=-9 \lambda+2 \lambda+10=0 \\\\ &=-7 \lambda=-10 \\\\ &\lambda=\frac{10}{7} \end{aligned}

so the value will be \frac{10}{7}

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