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  • Find the angle between the following pair of lines: (i) x - 2/2 = y- 1/5 = z- 3/- 3 and x+ 2/- 1 = y-4 /8 = z-5/ 4

11. Find the angle between the following pair of lines:

     (i)     \frac{x-2}{2}= \frac{y-1}{5}= \frac{z+3}{-3} and \frac{x+2}{-1}= \frac{y-4}{8}= \frac{z-5}{4}

Answers (1)

best_answer

Given lines are;

\frac{x-2}{2}= \frac{y-1}{5}= \frac{z+3}{-3}  and  \frac{x+2}{-1}= \frac{y-4}{8}= \frac{z-5}{4}

So, we two vectors \vec{b_{1}}\ and\ \vec{b_{2}} which are parallel to the pair of above lines respectively.

\vec{b_{1}}\ =2\widehat{i}+5\widehat{j}-3\widehat{k}   and  \vec{b_{2}}\ =-\widehat{i}+8\widehat{j}+4\widehat{k}

To find the angle A between the pair of lines \vec{b_{1}}\ and\ \vec{b_{2}} we have the formula;

\cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |

Then we have

\vec{b_{1}}.\vec{b_{2}} =(2\widehat{i}+5\widehat{j}-3\widehat{k}).(-\widehat{i}+8\widehat{j}+4\widehat{k})

=-2+40-12 = 26

and |\vec{b_{1}}| = \sqrt{2^2+5^2+(-3)^2} = \sqrt{38}

|\vec{b_{2}}| = \sqrt{(-1)^2+(8)^2+(4)^2} = \sqrt{81} = 9

Therefore we have;

\cos A = \left | \frac{26}{\sqrt{38} \times9} \right | = \frac{26}{9\sqrt{38}}

or A = \cos^{-1} \left ( \frac{26}{9\sqrt{38}} \right )

 

Posted by

Divya Prakash Singh

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