# Thread: Vectors, Distance Between Plane and Point

1. ## Vectors, Distance Between Plane and Point

i) Find the distance from the origin to the plane 3x + 2y - z - 14 = 0.

distance from point and plane = | [(0,7,0) • (3,2,-1)] / [√14] |
distance from the point and plane = |14 / √14 |
distance from the point and plane = 3.74 units

ii) Find the distance from the point P(10,10,10) to the plane 3x + 2y - z - 14 = 0.

distance from the point and plane = | [(-10,-3,-10) • (3,2,-1)] / [√14] |
distance from the point and plane = |-26 / √14 |
distance from the point and plane = 6.95 units

Textbook Answer: √14 or 3.74 units

iii) Is P on the same side of the plane as the origin? Explain.

2. Originally Posted by Macleef
i) Find the distance from the origin to the plane 3x + 2y - z - 14 = 0.
...
ii) Find the distance from the point P(10,10,10) to the plane 3x + 2y - z - 14 = 0.

distance from the point and plane = | [(-10,-3,-10) • (3,2,-1)] / [√14] |
distance from the point and plane = |-26 / √14 |
distance from the point and plane = 6.95 units

Textbook Answer: √14 or 3.74 units

iii) Is P on the same side of the plane as the origin? Explain.
to iii)

1. If you have the equation of a plane in the form:

$\displaystyle Ax+By+Cz+D=0$

then you know that the normal vector of the plane is $\displaystyle \vec n = [A, B, C]$ with the length $\displaystyle |\vec n | = \sqrt{A^2+B^2+C^2}$

If you calculate $\displaystyle \frac{\vec n}{|\vec n|} ={\vec n}^0$ then you get a vector with the same direction as $\displaystyle \vec n$ with the length 1. (= unit vector)

2. Divide the equation of the plane by the length of the normal vector:

$\displaystyle \frac{Ax + BY+CZ+D}{\sqrt{A^2+B^2+C^2}}=0$

The summand $\displaystyle \frac{D}{\sqrt{A^2+B^2+C^2}}= d$ is the distance of the origin to the plane.

Therefore at i) the distance of the origin to the plane is $\displaystyle d = \frac{14}{\sqrt{14}} = \sqrt{14}$

3. If you have a point P with it's stationary vector $\displaystyle \vec p$ then the dotproduct $\displaystyle {\vec n}^0 \cdot \vec p$ is the vertical projection of $\displaystyle \vec p$ on $\displaystyle {\vec n}^0$. If you now subtract the distance of the origin to the plane from the dotproduct you'll get the distance of the point P to the plane. Because you have to subtract the distance of the origin to the plane it is absolutely necessary that the constant of the equation of the plane is negative.

4. If the point $\displaystyle P_d$ and the origin are on different sides of the plane then the distance of $\displaystyle P_d$ to the plane is positive.
If $\displaystyle P_s$ and the origin are on the same side of the plane then the distance of $\displaystyle P_s$ to the plane is negative.

5. So don't use the absolute values when calculating the distance of a point to a plane and you can easily state on which sides of the plane the point and the origin are.

3. Originally Posted by Macleef
...

ii) Find the distance from the point P(10,10,10) to the plane 3x + 2y - z - 14 = 0.

distance from the point and plane = | [(-10,-3,-10) • (3,2,-1)] / [√14] |
distance from the point and plane = |-26 / √14 |
distance from the point and plane = 6.95 units

Textbook Answer: √14 or 3.74 units

...
According to #2 of my previous post I calculate the distance of P to the plane like this:

$\displaystyle d(P, p) = \frac1{\sqrt{14}} \cdot [10, 10, 10] \cdot [3, 2, -1] -\frac{14}{\sqrt{14}} =\frac{30+20-10-14}{\sqrt{14}} = \frac{26}{\sqrt{14}}$

And since d > 0 the point P and the origin are on different sides of the plane.