Three Dimensional Geometry Question 354
Question: A plane meets the co-ordinate axes in $ A,B,C $ and $ (\alpha ,\beta ,\gamma ) $ is the centered of the triangle $ ABC $ . Then the equation of the plane is
[MP PET 2004]
Options:
A) $ \frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3 $
B) $ \frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1 $
C) $ \frac{3x}{\alpha }+\frac{3y}{\beta }+\frac{3z}{\gamma }=1 $
D) $ \alpha x+\beta y+\gamma z=1 $
Show Answer
Answer:
Correct Answer: A
Solution:
Let the co-ordinates of the points where the plane cuts the axes are (a, 0, 0), (0, b, 0), (0, 0, c). Since centroid is $ (\alpha ,\beta ,\gamma ), $ therefore $ a=3\alpha ,, $ $ b=3\beta ,c=3\gamma . $ Equation of the plane will be $ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 $
$ \Rightarrow \frac{x}{3\alpha }+\frac{y}{3\beta }+\frac{z}{3\gamma }=1\Rightarrow \frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3. $
BETA
