Question

Show that the lines $\frac{5-x}{-4}=\frac{y-7}{4}=\frac{z+3}{-5}$ and $\frac{x-8}{7}=\frac{2y-8}{2}=\frac{z-5}{3}$ are coplanar.

Answer 1

Given lines are $\frac{5-x}{-4}=\frac{y-7}{4}=\frac{z+3}{-5}$ and $\frac{x-8}{7}=\frac{2y-8}{2}=\frac{z-5}{3}$

$\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}$ --- (1)

$\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}$--- (2)

Let the equtaion of plane passing through $(5,7,-3)$ be

$a(x-5)+b(y-7)+c(z+3)=0$--- (3)

$4a+4b-5c=0$ Normal to the plane 3 is $\perp$ to (1)-- (4)

$7a+b+3c=0$ Normal to the plane 3 is $\perp$ to (2)-- (5)

On solving (4) and (5), we get

$\frac{a}{12+5}=\frac{b}{-35-12}=\frac{c}{4-28}\Rightarrow \frac{a}{17}=\frac{b}{-47}=\frac{c}{-24}$

Putting the value of a, b, c in (3), we get

$17(x-5)-47(y-7)-24(z+3)=0$

Let us check whether point (8, 4, 5) lies on (6)

$17(8-5)-47(4-7)-24(5+3)=0\Rightarrow 0=0$

So, (6) is the required plane on which lines (1) and (2) lie and hence (1) and (2) are coplanar.