Solve this following


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

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

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

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

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

Correct Option: , 2



$\mathrm{A} \equiv(\mathrm{a}, 0,0), \mathrm{B} \equiv(0, \mathrm{~b}, 0), \mathrm{C} \equiv(0,0, \mathrm{c})$

Centroid $\equiv\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}, \frac{\mathrm{c}}{3}\right)=(1,1,2)$

$a=3, b=3, c=6$

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

$2 x+2 y+z=6$

line $\perp$ to the plane (DR of line $=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ )


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