question_answer

The plane\[2x+3y+4z=1\]meets the coordinate axes in A, B, C. The centroid of the triangle ABC is:

A)  (2, 3, 4)               

B)         \[\left( \frac{1}{2},\frac{1}{3},\frac{1}{4} \right)\]           

C)         \[\left( \frac{1}{6},\frac{1}{9},\frac{1}{12} \right)\]         

D)         \[\left( \frac{1}{2},\frac{3}{3},\frac{3}{4} \right)\]

Correct Answer: C

Solution :

Key Idea: If the plane \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] meets the coordinate axes, then the centroid of triangle is \[\left( \frac{a}{b},\frac{b}{3},\frac{c}{3} \right).\] Given, plane\[~2x+3y+4z=1\] meets the coordinate axes A, B, and C.                              \[\Rightarrow \]               \[\frac{2x}{1}+\frac{3y}{1}+\frac{4z}{1}=1\]                 \[\Rightarrow \]               \[\frac{x}{\frac{1}{2}}+\frac{x}{\frac{1}{3}}+\frac{z}{\frac{1}{4}}=1\] Then, centroid of the triangle ABC is                 \[\left[ \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right]\]                 \[=\left( \frac{\frac{1}{2}+0+0}{3}.\frac{0+\frac{1}{3}+0}{3}.\frac{0+0+\frac{1}{4}}{3} \right)\]                                 \[=\left( \frac{1}{6},\frac{1}{9},\frac{1}{12} \right).\]