Subjective Type

Find the values of $$p$$ so the line $$\cfrac{1-x}{3}=\cfrac{7y-14}{2p}=\cfrac{z-3}{2}$$ and $$\cfrac{7-7x}{3p}=\cfrac{y-5}{1}=\cfrac{6-z}{5}$$ are at right angles.

Solution

The given equations can be written in the standard form as
$$\cfrac{x-1}{-3}=\cfrac

{ y-2 }{ \cfrac { 2p }{ 7 } } =\cfrac{z-3}{2}$$ and $$\cfrac { x-1 }{

\cfrac { -3p }{ 7 } } =\cfrac{y-5}{1}=\cfrac{z-6}{-5}$$
The direction ratios of the lines are $$-3, \cfrac{-3p}{7}, 2$$ and $$\cfrac {-3p}{7}, 1, -5$$ respectively.
Two lines with direction ratios $${a}_{1}, {b}_{1}, {c}_{1}$$ and $${a}_{2}, {b}_{2}, {c}_{2}$$, are perpendicular to each other, if $${a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0$$.
$$\therefore$$ $$(-3)\left( \cfrac { -3p }{ 7 } \right)+\left( \cfrac { -3p }{ 7 } \right) .(1)+(2).(-5)=0$$
$$\Rightarrow$$ $$\cfrac{9p}{7}+\cfrac{2p}{7}=10$$
$$\Rightarrow$$ $$11p=70$$
$$\Rightarrow p=\cfrac{70}{11}$$

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