question_answer

Find the equation of the plane passing through the points \[(-\,1,\,\,2,\,\,0),\] \[(2,\,\,2,\,\,-\,1),\] and parallel to the line      \[\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-\,1}.\]

Answer:

Let the equation of plane passing through the point \[(-1,\,\,2,\,\,0)\,\,\text{be}\,\,A(x+1)+B(y-2)+C(z-0)=0\] where, A, B and C are direction ratios of the normal to the plane. Since, the plane also passes through \[(2,\,\,2,\,\,-\,1).\] \[\therefore \]      \[A(2+1)+B(2-2)+C(-\,1-\,0)=0\] \[\Rightarrow \]   \[3A+0\cdot B-C=0\]                  ?(ii) Given that, the plane is parallel to the line \[\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-\,1}\] or \[\frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-\,1}\] Therefore, we have             \[A+B-C=0\]                             ?(iii) [\[\because \] normal to the plane will be perpendicular to the line, and hence \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\]] On solving Eqs. (ii) and (iii), we get \[\frac{A}{0+1}=\frac{-\,B}{-\,3+1}=\frac{C}{3-0}\] \[\Rightarrow \]   \[\frac{A}{1}=\frac{-\,B}{-\,2}=\frac{C}{3}\] \[\Rightarrow \]   \[\frac{A}{1}=\frac{B}{2}=\frac{C}{3}=\lambda (say)\] On substituting the values of A, B and C in Eq. (i), we get \[1(x\,+1)+2(y\,-2)+3z=0\] \[\Rightarrow \]   \[x+1+2y-4+3z=0\] \[\therefore \]      \[x+2y+3z=3\] which is the required equation of plane.