A) \[bcx+cay+abz=0\]
B) \[bcx+cay-abz=0\]
C) \[bcx-cay+abz=0\]
D) \[-bcx+cay+abz=0\]
E) \[ax+by+cz=0\]
Correct Answer: B
Solution :
PA, PB are perpendiculars drawn from P(a, b, c) on\[yz\]and\[zx-\]planes. \[\therefore \] A(0, b, c) and B (a, 0, c) are points on\[yz\]and \[zx-\]planes. The equation of plane passing through (0, 0, 0) is \[Ax+By+Cz=0\] Which also passes through A and B. \[\therefore \] \[A.0+B.b+C.c=0\] ...(i) and \[A.a+B.O+C.c=0\] ...(ii) On solving Eqs. (i) and (ii), we get \[\frac{A}{bc-0}=\frac{B}{ac-0}=\frac{C}{0-ab}=\lambda (say)\] \[\Rightarrow \] \[A=\lambda bc,B=\lambda ac,C=-\lambda ab\] \[\therefore \]Required equation is \[bcx+acy-abz=0\]
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