Length of Perpendicular from a Given Point on a Given Plane

Let $P(x_1,y_1,z_1)$ be the given point. We have to find the length of perpendicular from this point to a plane. Let the equation of the plane in the normal form be \[lx+my+nz=p\text{ __(1)}\]

Consider another plane parallel to the given plane where the point $P(x_1,y_1,z_1)$ lies. The equation of the plane parallel to the given plane is \[lx+my+nz=p’\]

Since, it passes through $P(x_1,y_1,z_1)$, we have, \[lx_1+my_1+nz_1=p’\text{ __(2)}\]

Length of perpendicular from a given point on a given plane and distance between two planes

Let $O$ be the origin, then the distance between the two planes $\text{(1)}$ and $\text{(2)}$ is the difference between $ON’$ and $ON$ i.e. $p’$ and $p$.

If $PM$ be the perpendicular from $P$ on the given plane, then \[PM=N’N=ON’-ON=p’-p\] \[\therefore PM=lx_1+my_1+nz_1-p\]

This is the required length of the perpendicular.

If the point $P$ and the origin $O$ lie on the same side of the given plane, then it is observed that \[PM=N’N=ON-ON’=p-p’\] \[\therefore PM=p-(lx_1+my_1+nz_1)\] Thus, \[PM=lx_1+my_1+nz_1-p\]\[\text{or,}\:\;PM=p-lx_1-my_1-nz_1\] according as the points $P$ and $O$ lie on the opposite side or same side of the given plane.

Cor. Let the general equation of the given plane be \[Ax+By+Cz+D=0\] Reducing the given equation of the plane into the normal form, we have \[\frac{A}{\sqrt{\sum A^2}}x+\frac{B}{\sqrt{\sum A^2}}y+\frac{C}{\sqrt{\sum A^2}}z=\frac{-D}{\sqrt{\sum A^2}}\]

Hence the length of the perpendicular from the given point $P(x_1,y_1,z_1)$ will be \[\left|\frac{Ax_1+By_1+Cz_1+D}{\sqrt{\sum A^2}}\right|\]


Find the length of the perpendicular from the point $(2,5,7)$ to the plane $6x+6y+3z=11$.

Given point is $(2,5,7)$ and the plane is \[6x+6y+3z-11=0\] Length of the perpendicular from the given point to the given plane is \[\left|\frac{Ax_1+By_1+Cz_1+D}{\sqrt{A^2+B^2+C^2}}\right|\] \[\left|\frac{6×2+6×5+3×7-11}{\sqrt{6^2+6^2+3^2}}\right|\] \[=\left|\frac{52}{\sqrt{81}}\right|\] \[=\frac{52}{9}\:\;\text{units}\]


Find the distance between the planes $x-y+2z-4=0$ and $2x-2y+4z+5=0$.

Given equations of planes are \[x-y+2z-4=0\text{ __(1)}\] \[2x-2y+4z+5=0\text{ __(2)}\] Put $x=0$ and $y=0$ in plane $\text{(1)}$, we have \[2z=4\Rightarrow z=2\] Thus, $(0,0,2)$ is a point on the plane.

Length of the perpendicular from the point $(0,0,2)$ to the plane $\text{(2)}$ is \[\left|\frac{Ax_1+By_1+Cz_1+D}{\sqrt{\sum A^2}}\right|\] \[=\left|\frac{2×0-2×0+4×2+5}{\sqrt{2^2+2^2+4^2}}\right|\] \[=\left|\frac{13}{\sqrt{24}}\right|\] \[=\frac{13}{2\sqrt{6}}\:\;\text{units}\]


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