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  4. Equation of the plane perpendicular to the line (x/1) = (y/2) = (z/3) and Passing through the point (2, 3, 4) is

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Q. 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

KCETKCET 2014Three Dimensional Geometry

Solution:

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., $(a, b, c)=(1,2,3)$.
Now, from Eq. (i),
$1(x-2)+2(y-3)+3(z-4)= 0$
$\Rightarrow x+2 y+3 z= 20$