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

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

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

Answers (1)

best_answer

Given lines are;

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

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}+2\widehat{j}+\widehat{k}   and  \vec{b_{2}}\ =4\widehat{i}+\widehat{j}+8\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}+2\widehat{j}+\widehat{k}).(4\widehat{i}+\widehat{j}+8\widehat{k})

=8+2+8 = 18

and |\vec{b_{1}}| = \sqrt{2^2+2^2+1^2} = \sqrt{9} = 3

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

Therefore we have;

\cos A = \left | \frac{18}{ 3\times9} \right | = \frac{2}{3}

or A = \cos^{-1} \left ( \frac{2}{3} \right )

Posted by

Divya Prakash Singh

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