Question

Find the point in R^3 in the plane x +2y -3z =12 closest to the point (1,1,1)

Answer 1

Answer on Question #39341 - Math - Linear Algebra

Question. Find the point $B$ in $\mathbb{R}^3$ in the plane $p: x + 2y - 3z = 12$ closest to the point $A = (1,1,1)$.

Solution. The point $B$ in the plane $p$ closest to $A$ is a unique point on $p$ such that the vector $\overrightarrow{AB}$ is orthogonal to $p$. Thus to solve the problem we should find the intersection point of $p$ with the line $l$ orthogonal to $p$ and passing through $A$.

The assumption that $\overrightarrow{AB}$ is orthogonal to $p$ is equivalent to the assumption that $\overrightarrow{AB}$ is parallel to the normal vector $\vec{n}$ to $p$.

Coordinates of $\vec{n}$ are the coefficients of the left hand side of the equation of $p: x + 2y - 3z = 12$. Thus

$$\vec{n} = (1, 2, -3).$$

Therefore the parametric equation of the line $l$ orthogonal to $p$ and passing through $A$ is the following one:

$$\left\{ \begin{array}{l} x = 1 + t, \\ y = 1 + 2t, \\ z = 1 - 3t. \end{array} \right.$$

Substituting these formulas into the equation of $p$ we get:

$$1 + t + 2(1 + 2t) - 3(1 - 3t) = 12,$$ $$1 + t + 2 + 4t - 3 + 9t = 12,$$ $$14t = 12,$$ $$t = \frac{12}{14} = \frac{6}{7}.$$

Hence point $B$ has the following coordinates:

$$B = \left(1 + \frac{6}{7}, \; 1 + 2 \cdot \frac{6}{7}, \; 1 - 3 \cdot \frac{6}{7}\right) = \left(\frac{13}{7}, \; \frac{19}{7}, \; -\frac{11}{7}\right).$$

Answer.

$$B = \left(\frac{13}{7}, \; \frac{19}{7}, \; -\frac{11}{7}\right).$$