Question

Show that the lines $\frac{x-3}{2}=\frac{y+1}{-3}=\frac{z-2}{4}$ and $\frac{x+2}{2}=\frac{y-4}{4}=\frac{z+5}{2}$ are perpendicular to each other.

Answer 1

[To show: Lines are perpendicular to each other]

[i.e., To show: Angle between the given lines are ${90}^o$]

Lines: $\frac{x-3}{2}=\frac{y+1}{-3}+\frac{z-2}{4}$ and $\frac{x+2}{2}=\frac{y-4}{4}=\frac{z+5}{2}$

Angle between two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by

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

$a_1=2,b_1=-3,c_1=4$;

$a_2=2,b_2=4,c_2=2$

$\cos \theta =\frac{|4-12+8|}{\sqrt{4+9+16}\sqrt{4+16+4}}$

$\cos \theta =\frac{0}{\sqrt{29}\sqrt{24}}$

$\cos \theta =0$

$\Rightarrow \theta ={90}^o$

Hence, the given lines are perpendicular.