MHT CET · Maths · Three Dimensional Geometry
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
- 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 \(\left(\frac{3}{2}, \frac{3}{3}, \frac{3}{4}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{12}\right)\)
Step-by-step Solution
Detailed explanation
Note that
\(\mathrm{A} \equiv\left(\frac{1}{2}, 0,0\right), \mathrm{B} \equiv\left(0, \frac{1}{3}, 0\right), \mathrm{C} \equiv\left(0,0, \frac{1}{4}\right) \)
\( \therefore \text { Cendroid }=\left(\frac{\frac{1}{2}+0+0}{3}, \frac{0+\frac{1}{3}+0}{3}, \frac{0+0+\frac{1}{4}}{3}\right) \)
\( =\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{12}\right)\)
\(\mathrm{A} \equiv\left(\frac{1}{2}, 0,0\right), \mathrm{B} \equiv\left(0, \frac{1}{3}, 0\right), \mathrm{C} \equiv\left(0,0, \frac{1}{4}\right) \)
\( \therefore \text { Cendroid }=\left(\frac{\frac{1}{2}+0+0}{3}, \frac{0+\frac{1}{3}+0}{3}, \frac{0+0+\frac{1}{4}}{3}\right) \)
\( =\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{12}\right)\)
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