Question

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 perpendicular to each other. Find the value of $P$.

Answer 1

If the equation of two lines is given by:

$L_1:\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$

$L_2:\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$

Then, the angle between them is given by:

$\cos \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

Here, in our case:

$L_1:\frac{1-x}{3}=\frac{7y-14}{2P}=\frac{z-3}{2}$

$\Rightarrow L_1:\frac{x-1}{-3}=\frac{y-2}{2P/7}=\frac{z-3}{2}$

$L_2:\frac{7-7x}{3P}=\frac{y-5}{1}=\frac{6-z}{5}$

$\Rightarrow L_2:\frac{x-1}{-3P/7}=\frac{y-5}{1}=\frac{z-6}{-5}$

$a_1=-3,a_2=-\frac{3P}{7},b_1=\frac{2P}{7},b_2=1,c_1=2,c_2=-5$

Therefore:

$\cos \theta =\frac{|(-3\times -\frac{3P}{7})+(\frac{2P}{7}\times 1)+(2\times -5)|}{\sqrt{(-3{)}^2+{\left(\frac{2P}{7}\right)}^2+2^2}\sqrt{{(-\frac{3P}{7})}^2+1^2+(-5{)}^2}}$

$\Rightarrow \cos \theta =\frac{|\frac{9P}{7}+\frac{2P}{7}-10|}{\sqrt{13+\frac{4P^2}{49}}\sqrt{\frac{9P^2}{49}+26}}$

It is given that these lines are perpendicular, therefore $\theta =90˚\Rightarrow \cos \theta =0$

$⟹|\frac{9P}{7}+\frac{2P}{7}-10|=0$

$\Rightarrow \frac{11P}{7}=10$

$\Rightarrow P=\frac{70}{11}$