MHT CET · Maths · Three Dimensional Geometry
The Cartesian equation of a plane which passes through the points \(\mathrm{A}(2,2,2)\) and making equal nonzero intercepts on the co-ordinate axes is
- A \(x+y+z=6\)
- B \(x-2 y+z=0\)
- C \(2 x+y+z=7\)
- D \(x-y+z=6\)
Answer & Solution
Correct Answer
(A) \(x+y+z=6\)
Step-by-step Solution
Detailed explanation
Let \(a, b, c\) be the intercepts on the coordinate axes and we have \(a=b=c\).
\(\therefore \quad \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{a}}+\frac{\mathrm{z}}{\mathrm{a}}=1\) is required equation of plane.
Since plane passes through point \((2,2,2)\), we write \(\frac{2+2+2}{\mathrm{a}}=1 \Rightarrow \mathrm{a}=6\)
\(\therefore \quad\) Equation of plane is \(\mathrm{x}+\mathrm{y}+\mathrm{z}=6\)
\(\therefore \quad \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{a}}+\frac{\mathrm{z}}{\mathrm{a}}=1\) is required equation of plane.
Since plane passes through point \((2,2,2)\), we write \(\frac{2+2+2}{\mathrm{a}}=1 \Rightarrow \mathrm{a}=6\)
\(\therefore \quad\) Equation of plane is \(\mathrm{x}+\mathrm{y}+\mathrm{z}=6\)
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