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Prove that, In $R^{3}$ the distance $D$ between the point $P_{0}\left(x_{0}, y_{0}, z_{0}\right)$ and the plane $a x+b y+c z+d=0$ is $$D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$
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Proof

Let $Q\left(x_{1}, y_{1}, z_{1}\right)$ be any point in the plane, and let $\mathbf{n}=(a, b, c)$ be a normal vector to the plane that is positioned with its initial point at $Q .$ It is now evident that the distance $D$ between $P_{0}$ and the plane is simply the length (or norm) of the orthogonal projection of the vector $\overrightarrow{Q P}_{0}$ on $\mathbf{n}$,  is
$$D=\left\|\operatorname{proj}_{\mathbf{n}} \overrightarrow{Q P_{0}}\right\|=\frac{\left|\overrightarrow{Q P}_{0} \cdot \mathbf{n}\right|}{\|\mathbf{n}\|}$$
But
\begin{aligned} &\overrightarrow{Q P_{0}}=\left(x_{0}-x_{1}, y_{0}-y_{1}, z_{0}-z_{1}\right) \\ &\overrightarrow{Q P_{0}} \cdot \mathbf{n}=a\left(x_{0}-x_{1}\right)+b\left(y_{0}-y_{1}\right)+c\left(z_{0}-z_{1}\right) \\ &\|\mathbf{n}\|=\sqrt{a^{2}+b^{2}+c^{2}} \end{aligned}
Thus
$$D=\frac{\left|a\left(x_{0}-x_{1}\right)+b\left(y_{0}-y_{1}\right)+c\left(z_{0}-z_{1}\right)\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$
Since the point $Q\left(x_{1}, y_{1}, z_{1}\right)$ lies in the given plane, its coordinates satisfy the equation of that plane; thus
$$a x_{1}+b y_{1}+c z_{1}+d=0$$
or
$$d=-a x_{1}-b y_{1}-c z_{1}$$

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