Finding the distance(shortest) from a point to a plane

[Arnoldjavs3]

Homework Statement

assuming the point given is P(3,2,3) and the equation of the plane is 2x + y + 2z = 2, find the distance and the point Q which represents that point

Homework Equations

The Attempt at a Solution

Okay so I think that OQ-OP = QP, which connects the two points and is the line. This vector could be used to find the distance of the vector.

However, my book says that proj(n)(PoP) = QP as well. Why is this? n being the normal vector to the plane and Po being an arbritary point. What is the relation between the projection and this vector I'm describing? Since PoP is a vector on the plane, it should be orthogonal to the normal vector(in our case the normal vector can be (2,1,2)). I'm having a hard time visualizing the relation between the two

 

[Mark44]

Homework Statement

assuming the point given is P(3,2,3) and the equation of the plane is 2x + y + 2z = 2, find the distance and the point Q which represents that point

Homework Equations

The Attempt at a Solution

Okay so I think that OQ-OP = QP, which connects the two points and is the line. This vector could be used to find the distance of the vector.

However, my book says that proj(n)(PoP) = QP as well. Why is this? n being the normal vector to the plane and Po being an arbritary point. What is the relation between the projection and this vector I'm describing? Since PoP is a vector on the plane

No, it isn't. Presumably P₀ is a point in the plane, but P, as described above, is NOT a point in the plane. You can confirm this by substituting the coordinates of P into the plane's equation. The projection of P₀P in the direction of n (Proj_nP0P gives you the length of the line segment from the plane to the point P(3, 2, 3), with this line segment being perpendicular to the plane.

, it should be orthogonal to the normal vector(in our case the normal vector can be (2,1,2)). I'm having a hard time visualizing the relation between the two

 

[Arnoldjavs3]

No, it isn't. Presumably P₀ is a point in the plane, but P, as described above, is NOT a point in the plane. You can confirm this by substituting the coordinates of P into the plane's equation. The projection of P₀P in the direction of n (Proj_nP0P gives you the length of the line segment from the plane to the point P(3, 2, 3), with this line segment being perpendicular to the plane.

So projecting PoP on to N gives a vector that is parallel to n essentially correct? And this vector represents QP?

 

[Mark44]

So projecting PoP on to N gives a vector that is parallel to n essentially correct? And this vector represents QP?

Yes. Two things to keep in mind.
1. The component or projection of P₀P onto N is denoted in some books (such as Steward) as $comp_\vec{N} \vec{P_0P}$. That's a scalar and is the length you're looking for.
2. The vector projection of P₀P onto N is denoted as $proj_\vec{N} \vec{P_0P}$ and is a vector. It points in the same direction as N and has a magnitude equal to the distance between P₀ and P.

 

[Ray Vickson]

Homework Statement

assuming the point given is P(3,2,3) and the equation of the plane is 2x + y + 2z = 2, find the distance and the point Q which represents that point

Homework Equations

The Attempt at a Solution

Okay so I think that OQ-OP = QP, which connects the two points and is the line. This vector could be used to find the distance of the vector.

However, my book says that proj(n)(PoP) = QP as well. Why is this? n being the normal vector to the plane and Po being an arbritary point. What is the relation between the projection and this vector I'm describing? Since PoP is a vector on the plane, it should be orthogonal to the normal vector(in our case the normal vector can be (2,1,2)). I'm having a hard time visualizing the relation between the two

Let's clarify the geometry of the situation with a 2-dimensional example (which you can easily draw). What is the shortest distance between the point $P=(1,3)$ and the line $y = 1 + x?$

The unit normal vector to the line is $\vec{n} = (-1,1).$ Any vector $\vec{r} = (x,y)$ can be decomposed as
$$\vec{r} = \vec{r}_{\|} +\vec{r}_{\perp} , $$
where
$$\vec{r}_{\|} = \vec{r} - \vec{n} \;\; \vec{r} \cdot \vec{n}, \; \vec{r}_{\perp} = \vec{n} \; \vec{r} \cdot \vec{n}.$$
Here, $\vec{r}_{\|}$ is the component of $\vec{r}$ parallel to the line and $\vec{r}_{\perp}$ is the component perpendicular to the line. Note, however, that the vector $\vec{r}_{\|}$ does not lie in the line; it is parallel to the line but 1 unit away from it in the $y$-direction. For that reason, the vector $\vec{r}_{\perp}$ is too long to join the point P to the line; it joins P to a point on the opposite side of the line. However, it is in the right direction, so we can get the point $P_0$ in the line that is closest to $P$ from $\vec{P} = \vec{P_0} -t \vec{n},$ and set the scalar $t$ so that $P$ lies on the line.