AP EAMCET · Maths · Three Dimensional Geometry
A variable plane passes through a fixed point \((\alpha, \beta, \gamma)\) and meets the coordinate axes in \(A, B\) and \(C\). Let \(P_1, P_2\) and \(P_3\) be the planes passing through \(A, B, C\) and parallel to the coordinate planes \(Y Z, Z X, X Y\) respectively. Then, the locus of the point of intersection of the planes \(P_1, P_2\) and \(P_3\) is
- A \(\alpha x+\beta y+\gamma z=1\)
- B \(\frac{\alpha}{x}+\frac{\beta}{y}+\frac{\gamma}{z}=1\)
- C \(\alpha x^2+\beta y^2+\gamma z^2=1\)
- D \(\alpha \beta x+\beta \gamma y+\alpha \gamma z=1\)
Answer & Solution
Correct Answer
(B) \(\frac{\alpha}{x}+\frac{\beta}{y}+\frac{\gamma}{z}=1\)
Step-by-step Solution
Detailed explanation
Let the point \(A(a, 0,0), B(0, b, 0)\) and \(C(0,0, c)\). So, point of intersection of planes \(P_1, P_2\) and \(P_3\) is \(P(a, b, c)\) Now, equation of plane \(A B C\) is \(\because\) plane Eq. (i) passes through the point \((\alpha, \beta, \gamma)\), so…
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