Question

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

Answer 1

Equation of the given lines can be written as standard form
$\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}=\frac{z-3}{2}and\frac{x-1}{-\frac{3p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}$
Direction ratios of these lines are respectively $-3,\frac{2p}{7},2and\frac{-3p}{7}1,-5$
Two lines with direction ratios $a_1,b_1,c_{1}and{a}_2,b_2,c_2$ are perpendicular to each other, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$.
$\therefore (-3)\left(\frac{-3p}{7}\right)+\left(\frac{2p}{7}\right)\left(1\right)+\left(2\right)(-5)=0\Rightarrow \frac{9p}{7}+\frac{2p}{7}-10=0\Rightarrow \frac{11p}{7}=10\Rightarrow p=\frac{70}{11}$
Thus, the value of $p=\frac{70}{11}$
Note For making a line in standard form be ensure that coefficient of variable should be non-negative and constant.