Drawing a normal from a point P to a 3d plane

Question

through the point $P$ you draw a normal to the plane $2x + y-2z + 1 = 0$. Determine the coordinates of the point where the normal intersects the plane that $P$ has the coordinates. $P=(-3 ,0,2)$

I am new to linear algebra and don't really know what to do here and what equation I should really use.

My initial solution was $2(-3+n)+n-2(2+n)+1=-6+2n+n-4-2n+1=-6+n-4+1=>-9+n=0 = n=9$

$(-3+9, 0+9,2+9) = (6,9,11)$

which is wrong and what the professor said is that I didn't go along the normal so if any one can give me the equation or a help resource that I can use.

Answer 1

The normal to the plane is $$\left(\begin{matrix}2\\1\\-2\end{matrix}\right)$$ as can be seen from the coefficients of the plane equation. The equation of the line in the same direction as the normal and which passes through $P$ has equation $$\underline{r}=\left(\begin{matrix}-3\\0\\2\end{matrix}\right)+\lambda\left(\begin{matrix}2\\1\\-2\end{matrix}\right)=\left(\begin{matrix}-3+2\lambda\\\lambda\\2-2\lambda\end{matrix}\right)$$

Solving simultaneously with the plane, we have $$2(-3+2\lambda)+\lambda-2(2-2\lambda)+1=0\implies\lambda=1$$

Plugging this value into the line equation gives the coordinates of the intersection point $(-1,1,0)$

Answer 2

Let the intersection point be $Q$. Then $Q = P + t n $ for some $t $. To determine $t$, find the dot product of the above equation with the normal vector $n$, then you will have,

$Q \cdot n = P \cdot n + t (n \cdot n) $

Now, if a known point $p_0$ is on the plane, then we know that $Q - p_0$ is perpendicular to $n$. With this in mind, we'll add and subtract $p_0$ from the left hand side of the above equation,

$(Q - p_0 + p_0) \cdot n = P \cdot n + t (n \cdot n )$

This simplifies to

$ (Q - p_0) \cdot n + p_0 \cdot n = P \cdot n + t (n \cdot n)$

The first term on the left hand side is zero because as mentioned above, $(Q - p_0)$ lies in the plane and is therefore perpendicular to the normal vector $n$. Hence, the equation finally becomes,

$ p_0 \cdot n = P \cdot n + t (n \cdot n ) $

Solving for $t$,

$t = \dfrac{ (p_0 - P) \cdot n }{ n \cdot n }$

Now we can compute $Q = P + t n$

In this problem $P = (-3,0,2) , n = (2, 1, -2)$ and we can pick $p_0 = (0, -1, 0)$

Hence,

$t = \dfrac{ ( (0, -1, 0) - (-3, 0, 2) ) \cdot (2, 1, -2) } { (2, 1, -2) \cdot (2, 1, -2) } $

Simplifying,

$ t = \dfrac{ (3, -1, -2) \cdot (2, 1, -2) }{9} = \dfrac{9}{9} = 1 $

Therfore,

$Q = (-3, 0, 2) + (1) (2, 1, -2) = (-1, 1, 0) $