The plane $2 x+3 y+4 z=1$ meets $X$-axis in $A$, Y -axis in B and Z -axis in C . Then the centroid of…
The plane $2 x+3 y+4 z=1$ meets $X$-axis in $A$, Y -axis in B and Z -axis in C . Then the centroid of $\triangle A B C$ is
- $(2,3,4)$
- $\left(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right)$
- $\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{12}\right)$
- $\left(\frac{3}{2}, \frac{3}{3}, \frac{3}{4}\right)$
Solution
Note that
$\begin{aligned}
& \mathrm{A} \equiv\left(\frac{1}{2}, 0,0\right), \mathrm{B} \equiv\left(0, \frac{1}{3}, 0\right), \mathrm{C} \equiv\left(0,0, \frac{1}{4}\right) \\
& \therefore \quad \text { Cendroid }=\left(\frac{\frac{1}{2}+0+0}{3}, \frac{0+\frac{1}{3}+0}{3}, \frac{0+0+\frac{1}{4}}{3}\right) \\
&=\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{12}\right)
\end{aligned}$
Asked in: MHT CET 2024 (11 May Shift 1)
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