Distance from Origin to Plane
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Distance from Origin to Plane
A plane is a twodimensional doubly ruled surface spanned by two linearly
independent vectors. The generalization of the plane to higher dimensions is
called a hyperplane. The angle between two intersecting planes is known as the
dihedral angle.
The equation of a plane in 3D space is defined with normal vector (perpendicular
to the plane) and a known point on the plane.
Let the normal vector of a plane, and the known point on the plane, P1. And, let
any point on the plane as P.
We can define a vector connecting from P1 to P, which is lying on the plane.
Since the vector and the normal vector are perpendicular each other, the dot
product of two vector should be 0.
This dot product of the normal vector and a vector on the plane becomes the
equation of the plane. By calculating the dot product, we get;
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Distance from Origin
If the normal vector is normalized (unit length), then the constant term of the
plane equation, d becomes the distance from the origin.
If the unit normal vector (a1, b1, c1), then, the point P1 on the plane becomes
(Da1, Db1, Dc1), where D is the distance from the origin. The equation of the
plane can be rewritten with the unit vector and the point on the plane in order to
show the distance D is the constant term of the equation;
Therefore, we can find the distance from the origin by dividing the standard plane
equation by the length (norm) of the normal vector (normalizing the plane
equation). For example, the distance from the origin for the following plane
equation with normal (1, 2, 2) is 2;
Distance from a Point
The shortest distance from an arbitrary point P2 to a plane can be calculated by
the dot product of two vectors and , projecting the vector to the normal vector of
the plane.
where vP1 is a point on the plane and vNormal is the normal to the plane. I'm
curious as to how this gets you the distance from the world origin since the result
will always be 0. In addition, just to be clear (since I'm still kind of hazy on the d
part of a plane equation), is d in a plane equation the distance from a line through
the world origin to the plane's origin?
a) x+2y+3z=12
b) x+3y2z=4
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I'm not completely sure how to do this. Currently I'm plugging everything into the
equation:
d=abs(ax0+by0+cz0+d)/sqrt(a^2+b^2+c^2)
But I'm not sure if that's actually minimizing the distance.
Thanks in advance!
The shortest distance is along the perpendicular, i.e. the normal vector: <1,10,7>.
The line containing the origin:
r(t) = <1,10,7>t
The plane:
<(x  7), (y  5), (z + 8)> * <1,10,7> = 0
x  7 + 10(y  5) + 7(z + 8) = 0
Plug in the parametric equations from the line:
t  7 + 10(10t  5) + 7(7t + 8) = 0
... solve for t.
t = 1/150
Find the point using the line:
r(1/150) = (1/150, 1/15, 7/150)
Use the Distance Formula:
D = (1 + 10 + 7)/150
D = (1 + 100 + 49)/150
D = (150)/150
D = 1/150
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