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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 4 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 4 Maths Textbook Solution.

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Answer: the angle between the pairs of lines will be \frac{\lambda }{2}

Given: find the angle between the points of lines

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

Hint:\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}}}

Solution:\frac{x-2}{3}=\frac{y-(-3)}{-2}=\frac{z-5}{0} \\

              \vec{A}=3 \hat{i}-2 \hat{j}+0 \hat{k}

       Secondly: \frac{x-(-1)}{1}=\frac{y-3 / 2}{3 / 2}=\frac{z-5}{2}

       \cos \theta=\frac{(3)-3+0}{\sqrt{3^{2}+(-2)^{2}} \sqrt{1^{2}+(3 / 2)^{2}+(2)^{2}}} \\

        \cos \theta=0

       \cos \theta=\cos \frac{\lambda}{2}

       \theta=\frac{\pi}{2}

So the angle will be \theta =\frac{\lambda }{2}

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