Question

The equation of the plane passing through the origin and containing the lines whose d.c.'s are proportional to 1, -2, 2 and 2, 3, -1 is

• A. x - 2y + 2z = 0
• B. 2x + 3y - z = 0
• C. x + 5y - 3z =0
• D. 4x - 5y - 7z = 0

Answer 1

Answer: (D)

Any plane through the origin is $a(x-0)+b(y-0)+c(z-0)$ = 0.
The plane contains the lines whose d.c. are proportional to 1, - 2, 2 and 2, 3, -1 $\therefore$ normal to the plane are to $\perp$ to these lines.
$\therefore$ $a-2b+2c$ = 0
and $2a+3b+c$ = 0
$\therefore$ $\frac{a}{2-6}=\frac{b}{4+1}=\frac{c}{3+4}\Rightarrow \frac{a}{-4}=\frac{b}{5}=\frac{c}{7}$
$\therefore$ reqd . plane is
$-4\left(x-0\right)+5\left(y-0\right)+7\left(z-0\right)=0$
$\Rightarrow -4x+5y+7z=0$
$i.e.,4x-5y-7z=0$