Question

The equation of the plane passing through the origin and containing the line $\frac{x-1}{5}=\frac{y-2}{4}=\frac{z-3}{5}$ is

• A. $x+5y-3z=0$
• B. $x-5y+3z=0$
• C. $x-5y-3z=0$
• D. $3x-10y+5z=0$
• E. $x+5y+3z=0$

Answer 1

Answer: (B)

The equation of the plane through given line is $A(x-1)+B(y-2)+C(z-3)=0$ ...(i) where A, B and C are the DRs of the normal to the plane. Since, the straight line lie on the plane. .. DRs of the plane is perpendicular to the line, ie,
$5A+4B+5C=0$ ...(ii)
Since, it passes through (0, 0, 0), we get $-A-2B-3C=0$
$\Rightarrow$ $A+2B+3C=0$ ...(iii)
On solving Eqs. (ii) and (iii), we get
$\frac{A}{2}=\frac{B}{-10}=\frac{C}{6}$ From Eq. (i), $2(x-1)-10(y-2)+6(z-3)=0$
$\Rightarrow$ $2x-2-10y+20+6z-18=0$
$\Rightarrow$ $2x-10y+6z=0$
$\Rightarrow$ $x-5y+3z=0$