Question

The plane ax + by + cz = 1 meets the coordinate axes in A,B,C. The centroid if the triangle formed is
A. \[(3a,3b,3c)\]
B. \[(\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3})\]
C. \[(\dfrac{3}{a},\dfrac{3}{b},\dfrac{3}{c})\]
D. \[(\dfrac{1}{{3a}},\dfrac{1}{{3b}},\dfrac{1}{{3c}})\]

Answer 1

Hint: We take three coordinate axes X, Y and Z simultaneously. A point which lies on X-axis will have Y, Z coordinates zero, a point which lies on Y-axis will have X,Z coordinates zero and a point which lies on Z-axis will have X,Y coordinates zero. We take one point on each axis which joins to form a triangle. Then using the formula for centroid of a triangle we calculate the value.

Complete step-by-step answer:
We are given the plane \[ax + by + cz = 1\].
We take points on each axis separately where the plane meets the axis.
Let the plane meet the point on x-axis , then both the coordinates of $y$-axis and $z$-axis will be zero.
Since the plane meets the $x$-axis at point \[(x,0,0)\], the point satisfies the equation of the plane.
Therefore substituting the values \[y = 0,z = 0\] in the equation of plane \[ax + by + cz = 1\]
\[
\Rightarrow ax + b(0) + c(0) = 1 \\
\Rightarrow ax = 1 \\
 \]
Dividing both sides of the equation by \[a\]
\[
\Rightarrow \dfrac{{ax}}{a} = \dfrac{1}{a} \\
\Rightarrow x = \dfrac{1}{a} \\
 \]
Let the plane meet the point on the $y$-axis , then both the coordinates of $x$-axis and $z$-axis will be zero.
Since the plane meets the $y$-axis at point \[(0,y,0)\], so the point satisfies the equation of the plane.
\[
\Rightarrow a(0) + by + c(0) = 1 \\
by = 1 \\
 \]
Dividing both sides of the equation by \[b\]
\[
\Rightarrow \dfrac{{by}}{b} = \dfrac{1}{b} \\
\Rightarrow y = \dfrac{1}{b} \\
 \]
Let the plane meet the point on $z$-axis, then both the coordinates of $x$-axis and y-axis will be zero.
Since the plane meets the z-axis at point \[(0,0,z)\], the point satisfies the equation of the plane.
\[
\Rightarrow a(0) + b(0) + cz = 1 \\
\Rightarrow cz = 1 \\
 \]
Dividing both sides of the equation by \[c\]
\[
\Rightarrow \dfrac{{cz}}{c} = \dfrac{1}{c} \\
\Rightarrow z = \dfrac{1}{c} \\
 \]
Therefore we have three points on the plane
\[A(x,0,0),B(0,y,0),C(0,0,z)\]
Substituting the values of \[x = \dfrac{1}{a},y = \dfrac{1}{b},z = \dfrac{1}{c}\]
\[A(\dfrac{1}{a},0,0),B(0,\dfrac{1}{b},0),C(0,0,\dfrac{1}{c})\] and joining these three points will form a triangle \[\vartriangle ABC\]

Now we know centroid is a point which is exactly at the center of the triangle and using the formula for centroid \[\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3} + \dfrac{{{y_1} + {y_2} + {y_3}}}{3} + \dfrac{{{z_1} + {z_2} + {z_3}}}{3}} \right)\]
Where \[({x_1},{x_2},{x_3}),({y_1},{y_2},{y_3}),({z_1},{z_2},{z_3})\] are vertices of the triangle.
On comparing the vertices \[A,B,C\] we get
$\Rightarrow$\[({x_1} = \dfrac{1}{a},{x_2} = 0,{x_3} = 0),({y_1} = 0,{y_2} = \dfrac{1}{b},{y_3} = 0),({z_1} = 0,{z_2} = 0,{z_3} = \dfrac{1}{c})\]

$\therefore$ substituting values in formula for centroid of a circle
\[\left( {\dfrac{{\dfrac{1}{a} + 0 + 0}}{3} + \dfrac{{0 + \dfrac{1}{b} + 0}}{3} + \dfrac{{0 + 0 + \dfrac{1}{c}}}{3}} \right) = \left( {\dfrac{1}{{3a}},\dfrac{1}{{3b}},\dfrac{1}{{3c}}} \right)\]
So, option D is correct.

Note:
Students many times get confused with the word coordinate axis, they should know coordinate axis means the dimensions of that figure, here it is a three dimensional plane so we take three coordinate axis and a point on one axis will give all other coordinates of that point as zero.