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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 3 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 3 Maths Textbook Solution.

Answers (1)

Answer: the angle between the lines given in the question is \cos ^{-1}\frac{11}{14}

Given: find the angle between the pairs of lines;

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

Hint:

            \begin{aligned} &\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \end{aligned}

Solution:

                $$ \begin{aligned} &\frac{-(x-5)}{-2}=\frac{y-(-3)}{1}=\frac{-(3-1)}{3} \\ &\frac{x-b}{2}=\frac{y+3}{1}=\frac{(3+1)}{3} \end{aligned}

              Direction \: \: vector \vec{a}=2 \hat{i}+\hat{j}-3 \hat{k}

Again:

               \frac{x-0}{3}=\frac{(y-1)}{ 2}=\frac{(z+5)}{-1}

           \begin{aligned} Direction \: \: vector \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k} \end{aligned}

        \begin{aligned} &\cos \theta=\frac{6+2+3}{\left|\sqrt{2^{2}+1^{2}+(-3)^{2}}\right| \sqrt{3^{2}+(-2)^{2}+(-1)^{2}} \mid} \\ \end{aligned}

                            \frac{11}{\sqrt{14} \sqrt{14}}=\frac{11}{14} \\

                            \theta=\cos ^{-1} \frac{11}{14}

                           So the angle between the lines will be\cos ^{-1}\frac{11}{14}

                           

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