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

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.

• Are $\alpha,\beta$ your parameters in $\mathbb R$? Then $\bar{a},\bar{b}$ do not necessarily belong in $E$. For the pairs $(\alpha,\beta)=(1,0)$ and $(\alpha,\beta)=(0,1)$ you get that $A+ \bar{a},A+ \bar{b}$ are in $E$.
– T.P.
Nov 29, 2022 at 22:21

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 $$d(P,E) = d(P',E)$$.