The value of p so that the lines $$1$$−$$x3=7y$$−$$142p=z$$−$$32$$ and $$7$$−$$7x3p=y$$−$$51=6$$−$$z5$$ are at right angles are

x−$$1$$−$$3=y$$−$$22p7=z$$−$$32---$$ (1) x−$$1$$−$$3p7=y$$−$$51=z$$−$$6$$−$$5$$ $$---$$ (2) (1) and (2) are perpendicular. $$a1a2+b1b2+c1c2=0$$

Straight lines in 3D

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Question

The value of p so that the lines $\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2}$ and $\frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}$ are at right angles are

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Solution

$\frac{x - 1}{- 3} = \frac{y - 2}{\frac{2 p}{7}} = \frac{z - 3}{2}$ --- (1)
$\frac{x - 1}{\frac{- 3 p}{7}} = \frac{y - 5}{1} = \frac{z - 6}{- 5}$ --- (2)
(1) and (2) are perpendicular.
$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

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