question_answer

Find the value of \[\lambda \] such that the line \[\frac{x-2}{12}=\frac{y-1}{\lambda }=\frac{z-3}{-\,8}\] is perpendicular to the plane \[3x-y-2z=7.\]

Answer:

Given equation of line is \[\frac{x-2}{12}=\frac{y-1}{\lambda }=\frac{z-3}{-\,8}\]                                                                ?(i) and equation of plane is \[3x-y-2z=7\]       ...(ii) On comparing Eq. (i) with             \[\frac{x-{{x}_{1}}}{{{a}_{1}}}=\frac{y-{{y}_{1}}}{{{b}_{1}}}=\frac{z-{{z}_{1}}}{{{c}_{1}}},\] we get             \[{{a}_{1}}=12,\] \[{{b}_{1}}=\lambda \] and  \[{{c}_{1}}=-\,8\] On comparing Eq. (ii) with \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}},\] we get             \[{{a}_{2}}=3,\] \[{{b}_{2}}=-\,1,\] \[{{c}_{2}}=-\,2\] Since, the given line is perpendicular to given plane. So,       \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\] \[\Rightarrow \]   \[\frac{12}{3}=\frac{\lambda }{-\,1}=\frac{-\,8}{-\,2}\] \[\Rightarrow \]   \[4=-\,\lambda =4\] Hence,  \[\lambda =-\,4.\]