Question

The plane \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 3\] meets the co-ordinate axes in \[A,B,C\]. Then find the centroid of the triangle \[ABC\].
A. \[\left( {\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3}} \right)\]
B. \[\left( {\dfrac{3}{a},\dfrac{3}{b},\dfrac{3}{c}} \right)\]
C. \[\left( {\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}} \right)\]
D. \[\left( {a,b,c} \right)\]

Answer 1

Hint: Here, an equation of the plane and coordinates of the axes are given. First, simplify the equation of the plane in the intercept form. Then, calculate the intercepts of the plane on the axis. Then, apply the formula of the centroid of a triangle and calculate the required answer.

Formula used: The equation of the plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
The centroid of a triangle with vertices \[\left( {{x_1},{y_1},{z_1}} \right)\], \[\left( {{x_2},{y_2},{z_2}} \right)\], and \[\left( {{x_3},{y_3},{z_3}} \right)\] is: \[\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3},\dfrac{{{z_1} + {z_2} + {z_3}}}{3}} \right)\]

Complete step by step solution: Given:
The equation of the plane is \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 3\].
The plane meets the co-ordinate axes in \[A,B,C\].

Let’s simplify the equation of the plane in the intercept form.
Divide both sides by 3.
\[\dfrac{x}{{3a}} + \dfrac{y}{{3b}} + \dfrac{z}{{3c}} = \dfrac{3}{3}\]
\[ \Rightarrow \dfrac{x}{{3a}} + \dfrac{y}{{3b}} + \dfrac{z}{{3c}} = 1\]
From the above equation, we get
Co-ordinates of the point \[A\]: \[\left( {3a,0,0} \right)\]
Co-ordinates of the point \[B\]: \[\left( {0,3b,0} \right)\]
Co-ordinates of the point \[C\]: \[\left( {0,0,3c} \right)\]

Since \[ABC\] is a triangle and we have to calculate the co-ordinates of the centroid of the triangle.
So, apply the formula of the centroid of a triangle \[\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3},\dfrac{{{z_1} + {z_2} + {z_3}}}{3}} \right)\].
Substitute the values in the formula.
We get,
\[\left( {\dfrac{{3a + 0 + 0}}{3},\dfrac{{0 + 3b + 0}}{3},\dfrac{{0 + 0 + 3c}}{3}} \right)\]
\[ \Rightarrow \left( {\dfrac{{3a}}{3},\dfrac{{3b}}{3},\dfrac{{3c}}{3}} \right)\]
\[ \Rightarrow \left( {a,b,c} \right)\]
Thus, the co-ordinates of the centroid of the triangle \[ABC\] are \[\left( {a,b,c} \right)\].

Thus, Option (D) is correct.

Note: Remember the following forms of the equation of a plane:
General equation of a plane: \[ax + by + cz + d = 0\] , where \[d \ne 0\]
Equation of plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
Equation of a plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with direction ratios \[\left( {a,b,c} \right)\]: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\]