Question

Equation of a plane through the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and parallel to a coordinate axis is

• A. $4y-3z+1=0$
• B. $2x-z+1=0$
• C. $3x-2y+1=0$
• D. $2x+3y+1=0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $3x-2y+1=0$
B $2x-z+1=0$
C $4y-3z+1=0$

General equation of a plane is

$ax+by+cz+d=0$.........where a,b,c,d are constants taking any value

Let us consider the plane parallel to x-axis.

Hence, the equation of the plane can be written as $ay+bz=1$.

The plane passes through $(1,2,3)$ and the normal to the plane is perpendicular to the vector along the line.

Hence,
$2a+3b=1$ and $3a+4b=0$

On solving we get, $a=-4$ and $b=3$.

Hence, $-4y+3z=1$ or $4y-3z+1=0$.
Let us consider the plane parallel to the y-axis.

Hence, the equation of the plane can be written as:

$cx+dz=1$

$c+3d=1$ and $2c+4d=0$.

$\Rightarrow c=-2$ and $d=1$

Hence, the equation of the plane is $2x-z+1=0$.
Similarly, we can find the equation for the plane parallel to the z-axis. The equation of the plane can be written as $ex+fy=1$.

After substituting the value of the point $(1,2,3)$ and using the information of the normal to the plane being perpendicular to the line, we get the equation of the plane as $3x-2y+1=0$.
Hence, all three options are correct.