Question

Find the angle between the pair of lines:
$straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space 5 space straight j with hat on top space plus space straight k with hat on top space plus space straight lambda space left parenthesis 3 space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 6 space straight k with hat on top right parenthesis$
and $straight r with rightwards arrow on top space equals space 7 space straight i with hat on top space minus space 6 space straight k with hat on top space plus space straight mu space left parenthesis straight i with hat on top space plus space 2 space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis$

Solution

The equations of two lines are

straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space 5 space straight j with hat on top space plus space straight k with hat on top space plus space straight lambda space left parenthesis 3 space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 6 space straight k with hat on top right parenthesis

and

straight r with rightwards arrow on top space equals space 7 space straight i with hat on top space minus space 6 space straight k with hat on top space plus space straight mu left parenthesis straight i with hat on top space plus space 2 space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis

therefore

straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 6 space straight k with hat on top comma space space space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 2 space straight k with hat on top

Let θ be angle between the lines.

therefore space space space space cos space straight theta space equals space fraction numerator straight b with rightwards arrow on top. space stack straight b apostrophe with rightwards arrow on top over denominator open vertical bar straight b with rightwards arrow on top close vertical bar space open vertical bar straight b with rightwards arrow on top apostrophe close vertical bar end fraction space space space space space space space space space space space space space space space space space space space equals space fraction numerator left parenthesis 3 right parenthesis thin space left parenthesis 1 right parenthesis space plus space left parenthesis 2 right parenthesis space left parenthesis 2 right parenthesis space plus space left parenthesis 6 right parenthesis thin space left parenthesis 2 right parenthesis over denominator square root of 9 plus 4 plus 36 end root space square root of 1 plus 4 plus 4 end root end fraction space equals space fraction numerator 3 plus 4 plus 12 over denominator square root of 49 space square root of 9 end fraction space equals space fraction numerator 19 over denominator 7 cross times 3 end fraction space equals space 19 over 21 therefore space space space space space space space space straight theta space equals space cos to the power of negative 1 end exponent open parentheses 19 over 21 close parentheses

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