91Ó°ÊÓ

The given formula for distance from point$P\left(x_0,y_0,z_0\right)$ to plane$ax+by+cz+d=0$ is $\frac{k{x}_0+b{y}_0+c{z}_0+d}{\sqrt{a^2+b^2+c^2}}$.

The unit vector along the normal to the plane$\mathit{a}\mathit{x}\mathbf{+}\mathit{b}\mathit{y}\mathbf{+}\mathit{c}\mathit{z}\mathbf{+}\mathit{d}\mathbf{=}\mathbf{0}$ is given by $\hat{\mathbf{n}}\mathbf{=}\frac{\mathbf{a}\mathbf{i}\mathbf{+}\mathbf{b}\mathbf{j}\mathbf{+}\mathbf{c}\mathbf{k}}{\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}^{\mathbf{2}}}}$.

The given plane is ax + by + cz + d = 0. Let $\mathrm{P}\left({\mathrm{x}}_0,{\mathrm{y}}_0,{\mathrm{z}}_0\right)$be the point from which we calculate the distance to the plan D (say) and $B\left(x\text{'},y\text{'},z\text{'}\right)$. Further, let perpendicular from point P meets the plane at point A, i.e. AP is the normal to the plane.

Since, the point B$\left(x\text{'},y\text{'},z\text{'}\right)$lies on the plane, so

$ax\text{'}+by\text{'}+cz\text{'}+d=0...\left(1\right)$

The unit vector along the normal AN to the plane is given by (A is foot of perpendicular)$ $\hat{n}=\frac{ai+bj+ck}{\sqrt{a^2+b^2+c^2}}...\left(2\right)$

The vector $\stackrel{→}{\mathrm{PB}}=\mathrm{P}.\mathrm{V}$of point B-P.V of point P$=\left(x\text{'}i+y\text{'}i+z\text{'}i\right)-\left(x_{0}i+y_{0}j+z_{0}k\right)$that is,

$\overline{\mathrm{P}}\stackrel{→}{\mathrm{B}}=\left(\left(x\text{'}-x_0\right)i+\left(y\text{'}-y_0\right)i+\left(z\text{'}-z_0\right)i\right)........\left(3\right)$

From the figure it is clear that the distance of point $P(x_0,y_0,z_0)$from the plane is

$\mathrm{D}=\mathrm{PA}=\mathrm{\pm Ê\mu þ³¦´Ç²õÎ\pm }=\overline{\mathrm{P}}\stackrel{→}{\mathrm{B}}.\hat{\mathrm{n}}$

Now,

localid="1659076678645" $$\begin{gathered} \overline{\mathrm{P}}\stackrel{→}{\mathrm{B}}.\hat{\mathrm{n}}=\left(\left(x\text{'}-x_0\right)i+\left(y\text{'}-y_0\right)i+\left(z\text{'}-z_0\right)i\right).-\frac{ai+bj+ck}{\sqrt{a^2+b^2+c^2}} \\ =-\frac{\left(x\text{'}-x_0\right)a+\left(y\text{'}-y_0\right)b+\left(z\text{'}-z_0\right)}{\sqrt{a^2+b^2+c^2}} \\ =\frac{\left(x_{0}a+y_{0}b+z0c\right)-\left(x\text{'}a+y\text{'}b+z\text{'}c\right)}{\sqrt{a^2+b^2+c^2}} \end{gathered}$$

But from equation (1)

$ax\text{'}+by\text{'}+cz\text{'}=-d$

Now substitute $ax\text{'}+by\text{'}+cz\text{'}=-d$in above result, we have

$D=\frac{\left(x_{0}a+y_{0}b+z_{0}c\right)-(-d)}{\sqrt{a^2+b^2+c^2}}=\left|\frac{x_{0}a+y_{0}b+z_{0}c+d}{\sqrt{a^2+b^2+c^2}}\right|$

Thus for distance from point $\mathrm{P}\left({\mathrm{x}}_0,{\mathrm{y}}_0,{\mathrm{z}}_0\right)$to plane $ax+by+cz+d=0$is .

$\left|\frac{a{x}_0+b{y}_0+cz0+d}{\sqrt{a^2+b^2+c^2}}\right|$

Hence proved