JEE Main & Advanced Mathematics Three Dimensional Geometry Foot of Perpendicular from a Point \[\mathbf{A(\alpha ,}\,\,\mathbf{\beta ,}\,\,\mathbf{\gamma )}\]to a given plane \[\mathbf{ax+by+cz+d=0}\].

If AP be the perpendicular from A to the given plane, then it is parallel to the normal, so that its equation is \[\frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z-\gamma }{c}=r\] , (Say)

Any point P on it is \[(ar+\alpha ,\,br+\beta ,\,cr+\gamma )\]. It lies on the given plane and we find the value of \[r\] and hence the point P.

(1) Perpendicular distance : The length of the perpendicular from the point \[P({{x}_{1}},\,{{y}_{1}},{{z}_{1}})\] to the plane \[ax+by+cz+d=0\] is \[\left| \,\frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\, \right|\].

Distance between two parallel planes \[Ax+By+Cz+{{D}_{1}}=0\] and \[Ax+By+Cz+{{D}_{2}}=0\] is \[\frac{{{D}_{2}}\tilde{\ }{{D}_{1}}}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}\].

(2) Position of two points w.r.t. a plane : Two points \[P({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}})\] and \[Q({{x}_{2}},\,{{y}_{2}},{{z}_{2}})\] lie on the same or opposite sides of a plane \[ax+by+cz+d=0\] according to \[a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d\] and \[a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d\] are of same or opposite signs. The plane divides the line joining the points P and Q externally or internally according to P and Q are lying on same or opposite sides of the plane.