A) \[\alpha x,\beta y,\gamma z=3\]
B) \[\alpha x,\beta y,\gamma z=1\]
C) \[\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3\]
D) \[\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1\]
Correct Answer: C
Solution :
| [c] Let us take a triangle ABC and their vertices A (a, 0, 0), B(0, b, 0) and C(0, 0, c) |
| Therefore the equation of plane is |
| \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] (i) |
| Now, given centroid of \[\Delta ABC\]is \[(\alpha ,\beta ,\gamma )\] As we know, centroid of \[\Delta ABC\] with vertices \[({{x}_{1}},{{y}_{1}},{{z}_{1}}),({{x}_{2}},{{y}_{2}},{{z}_{2}})\] and \[({{x}_{3}},{{y}_{3}},{{z}_{3}})\] is given by \[\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3}, \right)\] |
| \[\therefore \] By using this formula, we have |
| \[\frac{a+0+0}{3}=\alpha \Rightarrow a=3\alpha ,;\frac{0+b+b}{3}=\beta \] |
| \[\Rightarrow b=3\beta \] |
| And \[\frac{0+0+c}{3}=\gamma \Rightarrow c=3\gamma \] |
| Now, put the values of a, b, c, in equation (i), which gives |
| \[\frac{x}{3\alpha }+\frac{y}{3\beta }+\frac{z}{3\gamma }=1\]\[\therefore \]\[\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3\] |
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