Question

Equation of the plane perpendicular to the line $\frac {x}{1} = \frac {y}{2} = \frac {z}{3}$ and Passing through the point $(2, 3, 4)$ is

• A. $x + 2y + 3z = 9$
• B. $x + 2y + 3z = 20$
• C. $2x + 3y + z = 17$
• D. $3x + 2y + z = 16$

Answer 1

Answer: (B)

$x + 2y + 3z = 20$

Answer 2

The correct option is(B): x +2 y +3 z =20.

Equation of plane passing through $(2,3,4)$ is
$a(x-2)+b(y-3)+c(z-4)=0$...(i)
Since, above plane is perpendicular to the line.
$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$
Thus, normal to the plane is parallel to the line.
So, DR's of normal is $(1,2,3)$, i.e., $\quad(a, b, c)=(1,2,3)$.
Now, from E (i),
$1(x-2)+2(y-3)+3(z-4)= 0$
$\Rightarrow x+2 y+3 z= 20$