KCET · Maths · Three Dimensional Geometry
If a plane meets the coordinate axes at \(A, B\) and \(C\) in such a way that the centroid of \(\triangle A B C\) is at the point \((1,2,3)\), then the equation of the plane is
- A \(\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=1\)
- B \(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\)
- C \(\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=\frac{1}{3}\)
- D \(\frac{x}{1}-\frac{y}{2}+\frac{z}{3}=-1\)
Answer & Solution
Correct Answer
(B) \(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\)
Step-by-step Solution
Detailed explanation
Given, plane meets the coordinate axes at \(A, B\) and \(C\).
Let the plane meets the coordinate axes at the points \(A(a, 0,0), B(0, b, 0), C(0,0, c)\)
The equation of plane is
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1...(i)\)
Centroid formula says,
\(\begin{aligned}
&x=\frac{x_{1}+x_{2}+x_{3}}{3}, y=\frac{y_{1}+y_{2}+y_{3}}{3} \\
&z=\frac{z_{1}+z_{2}+z_{3}}{3}
\end{aligned}\)
Centroid of \(\triangle A B C=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)\)
Given centroid \(=(1,2,3)\)
\(\Rightarrow\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)\)
Therefore, \(a=3, b=6\) and \(c=9\)
Put the value of \(a, b\) and \(c\) in equation of plane (i), we get
\(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\)
Let the plane meets the coordinate axes at the points \(A(a, 0,0), B(0, b, 0), C(0,0, c)\)
The equation of plane is
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1...(i)\)
Centroid formula says,
\(\begin{aligned}
&x=\frac{x_{1}+x_{2}+x_{3}}{3}, y=\frac{y_{1}+y_{2}+y_{3}}{3} \\
&z=\frac{z_{1}+z_{2}+z_{3}}{3}
\end{aligned}\)
Centroid of \(\triangle A B C=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)\)
Given centroid \(=(1,2,3)\)
\(\Rightarrow\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)\)
Therefore, \(a=3, b=6\) and \(c=9\)
Put the value of \(a, b\) and \(c\) in equation of plane (i), we get
\(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\)
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