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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 1 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 1 Maths Textbook Solution.

Answers (1)

Answer: the angle between the given pairs of lines will be  \cos ^{-1}=\frac{8}{5\sqrt{3}}

Given: find the angle between the following pairs of lines:

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

Hints: \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}

Solution:

          Directon ratio = 3,5,4

          Direction vector \vec{a}=3\hat{i}+5\hat{j}+4\hat{k}

          Again DR=1,1,2

          Direction vector \vec{b}=\hat{i}+\hat{j}+2\hat{k}

\begin{aligned} \cos \theta &=\frac{3+5+8}{\left|\sqrt{3^{2}+5^{2}+4^{2}}\right| \mid \sqrt{1^{2}+1^{2}+2^{2}}} \mid \\ &=\frac{16}{|\sqrt{50}| \sqrt{6} \mid} \\ =& \frac{8}{5 \sqrt{3}} \\ & \theta=\cos ^{-1} \frac{8}{5 \sqrt{3}} \end{aligned}

So the answer will be \theta =\cos ^{-1}\frac{8}{5\sqrt{3}}

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