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Every other point $(x,y,z)$ on the plane also generates a linear equation in the coefficients of the plane equation. In order to add it to the above system without reducing the dimension of the solution set, it must be dependent on the other equations, i.e., it must be a linear combination of the other three.
It looks like given the planar equation Ax+By+Cz+D=0, -D is equal to the dot product of the plane's normal with itself multiplied by a scalar c? $\endgroup$ – brain56 Commented Jul 20, 2019 at 20:27
You do not have enough information to specify the exact value of D in your equation. I would leave the equation as$$ x+11y+3z+D=0$$ until a point on the plane is given. At this point we have a family of parallel planes.
Suppose we have the plane with equation $3x-7z=12$. How to find its normal vector? The plane with equation ...
Can you please explain to me how to get from a nonparametric equation of a plane like this: $$ x_1−2x_2+3x_3=6$$ to a parametric one. In this case the result is supposed to be $$ x_1 = 6-6t-6s$$ $$ x_2 = -3t$$ $$ x_3 = 2s$$ Many thanks.
Find the equation of the plane perpendicular to the vector $\vec{n}\space=(2,3,6)$ and which goes through the point $ A(1,5,3)$. (A cartesian and parametric equation). Also find the distance between the beginning of axis and this plane. I'm not really sure where to start. Any help would be appreciated.
Find equation of plane in R3 that passes through point (2,3,5) and contains the line with equation ...
Since $\hat{n}$ is orthogonal to the plane, we have $$\begin{bmatrix}a\\b\\c \end{bmatrix}^T\begin{bmatrix}x-x_0\\y-y_0\\z-z_0 \end{bmatrix}=0,$$ then we obtain that formula. Note that, if $(x_0,y_0,z_0)=(0,0,0)$, then this is a plane (a subspace of $\mathbb{R}^3$) passing through the origin; if not, this plane is an affine plane (not a ...
I can think of two methods to find the normal vector to the plane. Method 1: Since the plane is orthogonal to $8x-2y+6z=1$, then the normal vector of the plane should be orthogonal to $(8,-2,6)$. So one normal vector would be $(1,1,-1)$.
The equation of an object is a way of telling whether a point is part of an object -- if you substitute the coordinates of the point into the equation and the equation is true, then the point is on the object; if the equation is not true for that point, then the point is not on the object.