MHT CET · Maths · Three Dimensional Geometry
A variable plane passes through the fixed point \((3,2,1)\) and meets \(X, Y\) and \(Z\) axes at points \(A\), B and C respectively. A plane is drawn parallel to YZ - plane through A , a second plane is drawn parallel to ZX -plan through B, a third plane is drawn parallel to XY - plane through C . Then locus of the point of intersection of these three planes, is
- A \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{11}{6}\)
- B \(\frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1\)
- C \(\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1\)
- D \(x+y+z=6\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1\)
Step-by-step Solution
Detailed explanation
Let the plane be \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
If passes through \((3,2,1)\)
\(\therefore \quad \frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1\)
Now, coordinates of points A, B, C are (a, 0, 0), \((0, b, 0)\) and \((0,0, c)\) respectively.
\(\therefore \quad\) Equations of the planes passing through \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are \(x=\mathrm{a}, y=\mathrm{b}\) and \(\mathrm{z}=\mathrm{c}\) respectively.
\(\therefore \quad\) From equation (i), we get
Required locus is \(\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1\)
If passes through \((3,2,1)\)
\(\therefore \quad \frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1\)
Now, coordinates of points A, B, C are (a, 0, 0), \((0, b, 0)\) and \((0,0, c)\) respectively.
\(\therefore \quad\) Equations of the planes passing through \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are \(x=\mathrm{a}, y=\mathrm{b}\) and \(\mathrm{z}=\mathrm{c}\) respectively.
\(\therefore \quad\) From equation (i), we get
Required locus is \(\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1\)
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