question_answer

The equation of plane passing through a point \[A(2,-1,\,3)\] and parallel to the vectors \[\mathbf{a}=(3,\,0,-1)\] and \[\mathbf{b}=(-3,\,\,2,\,2)\] is                                                                       [Orissa JEE 2005]

A)            \[2x-3y+6z-25=0\]

B)            \[2x-3y+6z+25=0\]

C)            \[3x-2y+6z-25=0\]

D)            \[3x-2y+6z+25=0\]

Correct Answer: A

Solution :

           As plane is parallel to a given vector Þ Normal of plane must perpendicular to the given vectors. Given point to which plane passes through is (2, ?1,3).                    Let A, B, C are direction ratios of its normal.                 \ Equation of plane is, \[A(x-2)+B(y+1)+C(z-3)=0\]                                                                                                        ?..(i)                    Now normal to plane \[A\mathbf{i}+B\mathbf{j}+C\mathbf{k}\] is perpendicular to the given vectors \[\mathbf{a}=3\mathbf{i}+0\mathbf{j}-\mathbf{k}\] and \[\mathbf{b}=-3\mathbf{i}+2\mathbf{j}+2\mathbf{k}\]                    \  \[3A+0B-C=0\]                                                ?..(i)                        \[-3A+2B+2C=0\]                                             .....(ii)                    Solving (i) and (ii) we get, \[\frac{A}{2}=\frac{B}{-3}=\frac{C}{6}\]                    \Equation of plane be \[2(x-2)-3(y+1)+6(z-3)=0\]                    i.e., \[2x-3y+6z-25=0\].