JEE Main & Advanced Mathematics Three Dimensional Geometry Equation of Plane Through a Given Line

(1) If equation of the line is given in symmetrical form as \[\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}\], then equation of plane is \[a(x-{{x}_{1}})+b(y-{{y}_{1}})+c(z-{{z}_{1}})=0\]                             .....(i)

where \[a,\,\,b,\,\,c\] are given by \[al+bm+cn=0\]           .....(ii)

(2) If equations of line is given in general form as \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0={{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}\], then the equation of plane passing through these line is \[({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}})\] \[+\lambda ({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}})=0\].

(3) Equation of plane through a given line parallel to another line : Let the d.c.’s of the other line be \[{{l}_{2}},\,{{m}_{2}},\,{{n}_{2}}\]. Then, since the plane is parallel to the given line, normal is perpendicular.

\ \[a{{l}_{2}}+b{{m}_{2}}+c{{n}_{2}}=0\]                                                      ……(iii)

Hence, the plane from (i), (ii) and (iii) is \[\left| \,\begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}}  \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}}  \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}}  \\ \end{matrix}\, \right|=0\].