Distance from Origin to Plane Distance from Origin to Plane A plane is a two-dimensional 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+3y-2z=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 = √(1² + 10² + 7²)/150 D = √(1 + 100 + 49)/150 D = √(150)/150 D = 1/√150

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