## Distance of a point from a plane

Determine the distance of the point $$A=(5, 5, 3, 3)^T$$ from the plane passing through the origin and the points $$B=(8, -1, 1, -2)^T$$ and $$C=(4, -2, 2, -1)^T$$.

• #### Hint

Use Gram-Schmidt ortonormalization. The distance is the norm of the vector $$\mathbf y_3$$, where vectors $$\mathbf x_1$$, $$\mathbf x_2$$ and $$\mathbf x_3$$ corresponds to the points $$B$$, $$C$$ and $$A$$.

• #### Resolution

With a minor pre-computation: vectors $$\mathbf x_1=(8, -1, 1, -2)^T$$, $$\mathbf x_2=(4, -2, 2, -1)^T$$ are linearly independent and indeed $$\begin{pmatrix} 8 & -1 & 1 & -2 \\ 4 & -2 & 2 & -1 \\ \end{pmatrix} \sim \begin{pmatrix} 8 & -4 & 4 & -2 \\ 8 & -1 & 1 & -2 \\ \end{pmatrix} \sim \begin{pmatrix} 8 & -4 & 4 & -2 \\ 0 & 3 & -3 & 0 \\ \end{pmatrix} \sim \begin{pmatrix} 4 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\ \end{pmatrix}$$.

The last two vectors $$\mathbf x_1'$$, $$\mathbf x_2'$$ are perpendicular and generate the same plane as $$\mathbf x_1$$ and $$\mathbf x_2$$.

We now project $$\mathbf x_3=(5, 5, 3, 3)^T$$:

$$p(\mathbf x_3)=\frac{\langle \mathbf x_3|\mathbf x_1' \rangle}{||\mathbf x_1'||^2}\mathbf x_1'+\frac{\langle \mathbf x_3|\mathbf x_2' \rangle}{||\mathbf x_2'||^2}\mathbf x_2' =\frac{17}{17}(4, 0, 0, -1)^T+\frac{2}{2}(0, 1, -1, 0)^T=(4, 1, -1, -1)^T$$

(We have used the non-normalized basis, only orthogonal hence the denominator contains the norm $$||\mathbf x_i'||$$ twice: once due to the normalization of $$\mathbf x_i'$$ and in the second time due to the product with the normalized $$\mathbf x_i'$$.)

We get $$||A-p(A)||=||\mathbf x_3-p(\mathbf x_3)||=||\mathbf y_3||=\sqrt{1+16+16+16}=7$$.