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To find the shortest distance between point and plane, we use the formula d = |Ax o + By o + Cz o + D |/√(A 2 + B 2 + C 2), where (x o, y o, z o) is the given point and Ax + By + Cz + D = 0 is the equation of the given plane.
Here's a quick sketch of how to calculate the distance from a point $P=(x_1,y_1,z_1)$ to a plane determined by normal vector $\vc{N}=(A,B,C)$ and point $Q=(x_0,y_0,z_0)$. The equation for the plane determined by $\vc{N}$ and $Q$ is $A(x-x_0)+B(y-y_0) +C(z-z_0) = 0$, which we could write as $Ax+By+Cz+D=0$, where $D=-Ax_0-By_0-Cz_0$.
The shortest distance will be achieved along a line that is perpendicular to the plane. The normal vector to the plane can be read off the equation: since the plane is 2x + 2y + z = 0, the normal vector of the plane is (2, 2, 1).
The distance from a point to a plane is equal to length of the perpendicular lowered from a point on a plane. If A x + B y + C z + D = 0 is a plane equation, then distance from point M(M x , M y , M z ) to plane can be found using the following formula
In Euclidean space, the distance from a point to a plane is the distance between a given point and its orthogonal projection on the plane, the perpendicular distance to the nearest point on the plane.
1 lip 2024 · Given a plane ax+by+cz+d=0 (1) and a point x_0= (x_0,y_0,z_0), the normal vector to the plane is given by v= [a; b; c], (2) and a vector from the plane to the point is given by w=- [x-x_0; y-y_0; z-z_0].
We've solved for the shortest distance between an arbitrary point and an arbitrary plane. If you like, you can expand the convert all the vectors to scalars to get exactly Sal's formula: n= [a,b,c] p0= [x0,y0,z0] ( [a,b,c)]* [x0,y0,z0]-D)/| (a,b,c)|=|d|. (ax0+by0+cz0-D)/sqrt (a^2+b^2+c^2)=|d|.
