question_answer

Find the equation of the plane passing through   the   point \[(1,\,\,1,\,\,-1)\] and perpendicular to the planes \[x+2y+3z-7=0\] and \[2x-3y+4z=0.\]

Answer:

Let the DR's of the normal to the required plane be a, b and c. Since, the required plane passes through the point \[(1,\,\,1,\,\,-1).\] Then, its equation is             \[a\,(x-1)+b(y-1)+c(z+1)\]             ? (i) Also, plane (i) is perpendicular to each of the given planes                \[x+2y+3z-7=0\] and       \[2x-3y+4z=0\] \[\therefore \]      \[a+2b+3c=0\]                          ? (ii) and       \[2a-3b+4c=0\]              ? (iii)                                                             \[\left[ \begin{align}   & \because \,\,\text{If}\,\,\text{two}\,\,\text{planes}\,\,\text{are}\,\text{perpendicular,}\,\,\text{then} \\  & {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0 \\  & \text{where,}\,\,{{a}_{1,}}\,{{b}_{1}},\,{{c}_{1}}\,\,\text{and}\,\,{{a}_{2,}}\,{{b}_{2}},\,{{c}_{2}}\,\,\text{are}\,\,\text{the}\,\,\text{DR }\!\!'\!\!\text{ s} \\  & \text{of}\,\,\text{the}\,\,\text{normal}\,\,\text{to}\,\,\text{the}\,\,\text{planes} \\ \end{align} \right]\] Solving Eqs. (ii) and (iii) by cross-multiplication method, we get             \[\frac{a}{8+9}=\frac{b}{6-4}=\frac{c}{-\,3-4}\] \[\Rightarrow \]   \[\frac{a}{17}=\frac{b}{2}=\frac{c}{7}=\lambda \,\,(say)\] \[\Rightarrow \]   \[a=17\lambda ,\] \[b=2\lambda \] and \[c=-\,7\lambda \] On putting the values of a, b and c in Eq. (i), we get \[17\lambda (x-1)+2\lambda (y-1)+(-7\lambda )(z+1)=0\] \[\Rightarrow \]   \[\lambda [17x-17+2y-2-7z-7]=0\] \[\Rightarrow \]   \[\lambda [17x+2y-7z-26]=0\] \[\therefore \]      \[17x+2y-7z=26\]          \[[\because \,\,\,\lambda \ne 0]\] Which is the required equation of plane.