Anmath14's question

[HallsofIvy]

Given three points p=(2,1,3), q=(1,0,1) and r=(2,-1,1). Find the Cartesian equation of the plane containing those points.

Any plane (except one parallel to the xy-plane which would have every z component the same- and that doesn't happen here) can be written in the for z= Ax+ By+ C for constants A, B, and C.

Since the point (2, 1, 3) lies in the plane, x= 2, y= 1, z= 3 satisfy the equation: 3= 2A+ B+ C.
Since the point (1, 0, 1) lies in the plane, x= 1, y= 0, z= 1 satisfy the equation: 1= A+ C.
Since the point (2, -1, 1) lies in the plane, x= 2, y= -1, z= 1 satisfy the equation: 1= 2A- B+ C.
Solve those three equations for A, B, and C.

For example, I observe that the second equation does not involve y while the first and third equations involve B with opposite coefficients! Adding the first and second equations, 4= 4A+ 2C so we have 1= A+ C and 2= 2A+ C. Subtracting the first of those two equations from the second eliminates C: 1= A. Then C= 0 and 1= 2- B so B= 1.

The equation of the plane is z= x+ y.

Check: 3= 2+ 1, 1= 1+ 0, and 1= 2- 1.

[Guest]

Let [tex]\vec{a}[/tex] denote the vector from the origin [tex]O[/tex] to a point [tex]A[/tex]. Now, let [tex]\hat{n}[/tex] be the unit normal vector to the plane. Also, let the shortest distance from [tex]O[/tex] to the plane be [tex]d[/tex].

Now, for a point [tex]T[/tex] on the plane, we know that [tex]\vec{t}\cdot\hat{n}=d[/tex] because [tex]d[/tex] is the calculated as the length of the vector from [tex]O[/tex] the plane and the [tex]\vec{t}\cdot\hat{n}[/tex] gives the component of [tex]\vec{t}[/tex] along [tex]\hat{n}[/tex].

So, [tex]\vec{t}\cdot\hat{n}=d=\vec{p}\cdot\hat{n}\Rightarrow \vec{t}\cdot\vec{n}=\vec{p}\cdot\vec{n}[/tex]

In the above equation, replace [tex]\vec{t}[/tex] with [tex]\vec{t}=x\hat{i}+y\hat{j}+z\hat{k}[/tex], compute [tex]\vec{n}=\vec{qp}\times\vec{qr}[/tex].

So, [tex]\vec{qp}=\vec{p}-\vec{q}=\hat{i}+\hat{j}+2\hat{k}[/tex], [tex]\vec{qr}=\vec{r}-\vec{q}=\hat{i}-\hat{j}[/tex] and [tex]\vec{n}=\vec{qp}\times\vec{qr}=2\hat{i}+2\hat{j}-2\hat{k}[/tex].

Putting it in the equation above, [tex]\left(x\hat{i}+y\hat{j}+z\hat{k}\right)\cdot\left(2\hat{i}+2\hat{j}-2\hat{k}\right)=\left(2\hat{i}+\hat{j}+3\hat{k}\right)\cdot\left(2\hat{i}+2\hat{j}-2\hat{k}\right)\Rightarrow 2x+2y-2z=0\Rightarrow\boxed{x+y=z}[/tex]