Location of the point, which lies on the line and has the same distance to a given plane as another given point.

Question

Given a line g: $\bar x = P + \lambda \bar t$ and the plane $E$: $\bar x = A + \alpha \bar a + \beta \bar b $ . Determine the location of the point $P'\neq P$, which lies on g and has the same distance to the plane $E$ as the point $P$

$P=(1,0,0)$

$\bar t=(1,1,1)$

$A=(2,3,0)$

$\bar a=(2,0,0)$

$\bar b=(0,-1,1)$.

So I figured that vectors $\bar a$ and $\bar b$ should belong to $E$, hence their cross product will be normal to the plane, which I have found $\bar n=(0,-2,-2)$. Using $\bar n$ and the coordinates of point A I came up with the equation of the plane in different form $E=\bar n(\bar x - A)=0 \rightarrow 2x_2 -2x_3-6=0$ And then I got lost, the formula for distance between plane and point would give $3$ unknowns in one equation. Clearly, I should look into some other direction.

Answer 1

The normal vector is wrong. A normal of the plane is $\overline n =(0,1,1)$ and the equation becomes $x_2 + x_3 = 3$.

First, find the intersection of $g$ and $E$. Substituting $(1+ \lambda, \lambda, \lambda)$ into $x_2 + x_3 = 3$ gives $\lambda = 3/2$. Then, plug in the value $2\lambda = 3$ (the double of $\lambda)$ in the parametric equation of $g$, and you get $P'= (4,3,3)$.

Edit: I will give an intuitive explanation why we plug in $2\lambda$ to obtain $P'$. Have a look at the picture.

We have shown that for $\lambda = 3/2$, we get the intersection point, say $Q$, of the line and the plane. So, if it takes "3/2 steps" to cover the distance from $P$ to the plane, then we need to take another "3/2 steps" to arrive at the point $P'$.

Note that the distance $\|P-Q\|$ is not the distance from $P$ to the plane. The line $g$ is not normal to the plane $E$. If you want to calculate the distance from the point $P$ to the plane, you need the formula $$ d(P,E) = \frac{|(P-Q) \cdot \vec{n}|}{\|\vec{n}\|}. $$ (We take the inproduct of the vector $P-Q$ with the unit normal $\vec{n}/\|\vec{n}\|$.) You can check that $d(P,E) = d(P',E)$.