Three Dimensional Geometry Question 179
Question: The equation of the plane which makes with co-ordinate axes, a triangle with its centroid $ (\alpha ,\beta ,\gamma ) $ is
Options:
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 $
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Answer:
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 $ ( \frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3}, ) $
$ \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 $
BETA
