Question

If a plane meets coordinate axes in A, B, C such that the centroid of the triangle is \[\left( {1,k,{k^2}} \right)\] the equation of the plane is:-
A) \[x + ky + {k^2}z = 3{k^2}\]
B) \[{k^2}x + ky + z = 3{k^2}\]
C) \[x + kx + {k^2}z = 3\]
D) \[{k^2}x + ky + z = 3\]

Answer 1

Hint: Here first we will assume the equation of the plane to be
\[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]. Then we will find the centroid of the triangle formed and find the values of a, b and c and then substitute them in the equation of the plane to get the desired answer.

Complete step-by-step answer:
Let us assume the equation of the plane to be
\[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]……………………………………..(1)
Now since it is given that the plane meets the coordinate axes in A, B, C
Therefore the coordinates of A, B, C are \[A(a,0,0);B\left( {0,b,0} \right);C\left( {0,0,c} \right)\]
Now we know that the coordinates of centroid of triangle having vertices as \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\]are given by:-
\[x = \dfrac{{{x_1} + {x_2} + {x_3}}}{3};y = \dfrac{{{y_1} + {y_2} + {y_3}}}{3};z = \dfrac{{{z_1} + {z_2} + {z_3}}}{3}\]
We are given the centroid of the triangle as \[\left( {1,k,{k^2}} \right)\]
Hence applying the formula of centroid for above triangle we get:-
For x coordinate:
$\Rightarrow$\[1 = \dfrac{{a + 0 + 0}}{3}\]
Solving for a we get:-
$\Rightarrow$\[1 = \dfrac{a}{3}\]
\[ \Rightarrow a = 3\]
For y coordinate:-
\[k = \dfrac{{0 + b + 0}}{3}\]
Solving for b we get:-
$\Rightarrow$\[k = \dfrac{b}{3}\]
\[ \Rightarrow b = 3k\]
For z coordinate:-
$\Rightarrow$\[{k^2} = \dfrac{{0 + 0 + c}}{3}\]
Solving for c we get:-
$\Rightarrow$\[{k^2} = \dfrac{c}{3}\]
\[ \Rightarrow c = 3{k^2}\]
Now putting these values of a, b, c in equation 1 we get:-
$\Rightarrow$\[\dfrac{x}{3} + \dfrac{y}{{3k}} + \dfrac{z}{{3{k^2}}} = 1\]
Taking LCM we get:-
$\Rightarrow$\[\dfrac{{x\left( {{k^2}} \right) + y\left( k \right) + z\left( 1 \right)}}{{3{k^2}}} = 1\]
Solving it further we get:-
$\Rightarrow$\[\dfrac{{{k^2}x + ky + z}}{{3{k^2}}} = 1\]
Cross multiplying we get:-
$\Rightarrow$\[{k^2}x + ky + z = 3{k^2}\]
Hence the equation of the plane is \[{k^2}x + ky + z = 3{k^2}\]

Therefore, option B is correct.

Note: Students should take a note that the equation of the plane is to be taken in the form of
\[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\] in order to make the question easier. Otherwise, the students can also use the other forms of equations of a plane.
Centroid of a triangle is the intersection point of the medians drawn from each vertex of the triangle to its opposite side. The centroid divides the median into the ratio of 2:1