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

In summary: In this case, we would have ##x = 1 + t, y = 1 + t + 3t = 1 + 4t,## so that ##P = (1+4t, 1+4t).## Plugging this into the line's equation, we get$$1 + 4t = 1 + (1+4t),$$which is true for any ##t##. Therefore, the distance between ##P## and the line is just the magnitude of ##\vec{r}_{\perp},## which is$$|\vec{r}_{\perp}| = |2 \vec{n} \cdot \vec{r}| = 2 |
  • #1
Arnoldjavs3
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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
 
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  • #2
Arnoldjavs3 said:

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 (ProjnP0P 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.
Arnoldjavs3 said:
, 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
 
  • #3
Mark44 said:
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 (ProjnP0P 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?
 
  • #4
Arnoldjavs3 said:
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.
 
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  • #5
Arnoldjavs3 said:

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.
 

1. What is the formula for finding the distance from a point to a plane?

The formula for finding the distance from a point to a plane is: d = |ax + by + cz + d| / √(a^2 + b^2 + c^2), where (x, y, z) is the coordinates of the point and the plane is represented by the equation ax + by + cz + d = 0.

2. How do I determine if a point is above or below a plane?

To determine if a point is above or below a plane, you can substitute the coordinates of the point into the equation of the plane. If the result is positive, the point is above the plane. If the result is negative, the point is below the plane.

3. Can the shortest distance from a point to a plane be negative?

No, the shortest distance from a point to a plane cannot be negative. It is always measured as a positive value.

4. Is there a geometric interpretation of finding the distance from a point to a plane?

Yes, the distance from a point to a plane can be thought of as the length of the perpendicular line drawn from the point to the plane. This line is also known as the "normal" or "altitude" of the point to the plane.

5. Can the distance from a point to a plane be 0?

Yes, the distance from a point to a plane can be 0 if the point lies on the plane. This means that the point is coplanar with the plane and there is no distance between them.

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