Question

A plane $P$ meets the coordinate axes at $A,B$ and $C$ respectively. The centroid of a triangle $ABC$ is given to be $(1,1,2)$. Then the equation of the line through this centroid and perpendicular to the plane $P$ is:

• A. $\frac{x-1}{2}=\frac{y-1}{1}=\frac{z-2}{1}$
• B. $\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}$
• C. $\frac{x-1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$
• D. $\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-2}{2}$

Answer 1

The correct option is B

$\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}$

Explanation for the correct option:

Step 1. Draw the figure according to the question:

From the figure, we get

Centroid, $G(\frac{\alpha }{3},\frac{\beta }{3},\frac{\gamma }{3})=(1,1,2)$

$\Rightarrow$$\alpha =3,\beta =3,\gamma =6$

Step 2. Equation of plane is $\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$

$\Rightarrow$$\frac{x}{3}+\frac{y}{3}+\frac{z}{6}=1$

$\Rightarrow$ $2x+2y+z=6$

$\therefore$Required line perpendicular to plane $P$ is $\frac{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{}}{\mathbf{2}}\mathbf{=}\mathbf{}\frac{\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{2}}\mathbf{=}\mathbf{}\frac{\mathbf{(}\mathbf{z}\mathbf{-}\mathbf{2}\mathbf{)}}{\mathbf{1}}$

Hence, Option ‘B’ is Correct.