Question

If a plane meets the co-ordinate axes at \[A,B\] and \[C\] such that the centroid of the triangle is \[\left( {1,2,4} \right)\] , then what is the equation of the plane?
A. \[x + 2y + 4z = 12\]
B. \[4x + 2y + z = 12\]
C. \[x + 2y + 4z = 3\]
D. \[4x + 2y + z = 3\]
E. \[x + y + z = 12\]

Answer 1

Hint: Here, a centroid of a triangle is given. First, calculate the vertices of the triangle \[ABC\]. Then, apply the formula of the centroid of the triangle and calculate the intercepts of the plane. In the end, solve the equation of the plane in the intercept form and get 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:
A plane meets the co-ordinate axes at \[A,B\] and \[C\].
The centroid of the triangle \[ABC\] is \[\left( {1,2,4} \right)\].

Since the plane meets the co-ordinate axes at \[A,B\], and \[C\].
So, the co-ordinates of the points \[A,B\], and \[C\] are: \[A\left( {a,0,0} \right),B\left( {0,b,0} \right)\] and \[C\left( {0,0,c} \right)\]

It is given that the centroid of the triangle \[ABC\] is \[\left( {1,2,4} \right)\].
Apply the formula for the centroid of the triangle.
We get,
\[\left( {\dfrac{{a + 0 + 0}}{3},\dfrac{{0 + b + 0}}{3},\dfrac{{0 + 0 + c}}{3}} \right) = \left( {1,2,4} \right)\]
\[ \Rightarrow \left( {\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3}} \right) = \left( {1,2,4} \right)\]
\[ \Rightarrow a = 3,b = 6,c = 12\]
So, the intercepts of the plane are \[3,6\] and \[12\].

Now write the equation of the plane in the intercept form.
We get,
\[\dfrac{x}{3} + \dfrac{y}{6} + \dfrac{z}{{12}} = 1\]
\[ \Rightarrow \dfrac{{x \times 4}}{{3 \times 4}} + \dfrac{{y \times 2}}{{6 \times 2}} + \dfrac{z}{{12}} = 1\]
\[ \Rightarrow \dfrac{{4x}}{{12}} + \dfrac{{2y}}{{12}} + \dfrac{z}{{12}} = 1\]
\[ \Rightarrow \dfrac{{4x + 2y + z}}{{12}} = 1\]
\[ \Rightarrow 4x + 2y + z = 12\]
Thus, the equation of the required plane is \[4x + 2y + z = 12\].

Thus, Option (B) 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 \[\left( {a,b,c} \right)\]: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\]