Problem on finding a plane from three points
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Find the plane containing the points $(a,0,0)$, $(0, b, 0)$, and $(0,0,c)$.
Solution
Recall that:
A plane is specified by a point and a normal vector
To specify a plane, we need a point $\mathbf{x_0}$ and a normal vector $\mathbf{n}$.
The plane going through $\bfx_0$ with normal vector $\bfn$ is given by $\bfn \cdot \bfx = \bfn \cdot \bfx_0.$
Because the plane goes through $(a, 0, 0)$, we choose $\bfx_0 = (a, 0, 0)$.
Now, we seek a normal vector $\bfn$.
Because $\bfn$ is a normal vector, it is perpendicular to any vector along the plane.
In particular, it is perpendicular to the vectors from $(a,0,0)$ to $(0,b,0)$ and from $(a,0,0)$ to $(0,0,c)$.
Subtraction gives the vector between two points
These vectors are $(-a, b, 0)$ and $(-a, 0, c)$, respectively.
Recall that
Cross product gives a perpendicular vector
The cross product of two vectors is perpendicular to both.
Cross product definition
To get $\bfn$, we compute the cross product \begin{align}\bfn &= (-a, b, 0) \times (-a, 0, c) \\ &= (bc, ac, ab) \end{align}
Direct computation of dot product
We can compute $$ \bfn \cdot \bfx_0 = abc.$$
Hence, the plane is given as $(bc, ac, ab) \cdot \bfx = abc$.
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