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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 9 Sub Question 5 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 9 Sub Question 5 Maths Textbook Solution.

Answers (1)

Answer: the angle between the pairs of given line is \cos ^{-1}\frac{4}{5\sqrt{6}}

Given: find the angle between the pairs of lines

\begin{aligned} &\frac{x-5}{1}=\frac{2 y+6}{-2}=\frac{z-3}{1} \text { and } \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-6}{5} \\ \end{aligned}

Hint:\begin{aligned} &\cos \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}}} \\ \end{aligned}

Solution:

      \begin{aligned} &\frac{x-5}{1}=\frac{y+3}{-1}=\frac{z-3}{1} \\\\ &a_{1}, b_{1}, c_{1}=1,-1,1 \\ \end{aligned}

again,

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

      \cdot a_{2}, b_{2}, c_{2}=3,4,5 \\\\

\cos \theta=\frac{3-4+5}{\sqrt{1^{2}+(-1)^{2}+1^{2}} \sqrt{3^{2}+4^{2}+5^{2}}}

            \begin{aligned} &\quad=\frac{8-4}{\sqrt{3} \sqrt{50}}=\frac{4}{5 \sqrt{6}} \\ &\cos \theta=\cos ^{-1} \frac{4}{5 \sqrt{6}} \end{aligned}

So the angle will be\cos ^{-1}\frac{4}{5\sqrt{6}}

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