Line of Intersection of Planes (Closest Point)

Question

I found the line of intersection of two planes and found the parametric equations for that line of intersection.

But how do I find a point on that line of intersection that is closest to a certain point that is given to me in the problem?

For example my parametric equations I found for the line of intersection of the planes, 2x + 10y + 2z= -2 and 4x + 2y - 5z = -4 are

x=-2-6t y=2t z=-4t

and I need to find a point one the line of intersection that is closest to point (12,14,0).

Would I need to use lagrange multipliers?

Answer 1

No, you wouldn't need to use Lagrange multipliers.

Find $t$ that minimizes the square of the distance from $(-2-6t,2t,-4t)$ to $(12,14,0)$,

which is $(-2-6t-12)^2+(2t-14)^2+(-4t)^2=56(t+1)^2+336$,

which is minimized when $t=-1$.

Answer 2

If this exercise is within the context of analytical geometry, I would rather suggest the following approach:

Let $P = \begin{pmatrix}-2 \\ 0 \\ 0 \end{pmatrix}$, $d = \begin{pmatrix}-6 \\ 2 \\ -4 \end{pmatrix}$ and $Q = \begin{pmatrix}12 \\ 14 \\ 0 \end{pmatrix}$. Then the line is $P + td$ and $PQ = Q-P = \begin{pmatrix}14 \\ 14 \\ 0 \end{pmatrix}$.

You find the closest point to Q on the line by orthogonally projecting $PQ$ onto the line:

$$\left(PQ \cdot \frac{d}{|d|}\right)\frac{d}{|d|} = \frac{PQ \cdot d}{|d|^2}d =-d$$

Hence, $P-d = \begin{pmatrix}4 \\ -2 \\ 4 \end{pmatrix}$ is the closest point.