243.12.7.12 plane from 3 points

[karush]

$\tiny\textit{243.12.7.12}$

$\textsf{Write the equation for the plane through points}$
$$A(-3,8,1),\, B(5,-7,-18),\, C(-1,-5,24)$$
ok, tried to follow an example of cross product but..
$$\displaystyle\vec{AB}\times \vec{AC} =
\begin{bmatrix}
\textbf{i} & \textbf{j} & \textbf{k}\\
\, 5 &-7 &-1\\
-1 &-5 &-1
\end{bmatrix}=$$
book answer $\color{red}{ \displaystyle 8x + 3y + z = 1 }$

 

[Greg]

$$ax+by+cz=d$$

$$\vec{AB}\times \vec{AC} =
\begin{bmatrix}
\textbf{i} & \textbf{j} & \textbf{k}\\
\, 8 &-15 &-19\\
2 &-13 &23
\end{bmatrix}$$

Divide the coefficients of the resulting vector by $-74$, plug in point $A$ to find $d$, then check the result by plugging in points $B$ and $C$.

 

[karush]

sorry didn't really follow that

 

[Greg]

Can you be specific? Exactly what is it that you don't understand?

 

[HallsofIvy]

In your first post you wrote "\(\displaystyle \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 8 & -15 & -19 \\ 2 & -13 & 23\end{array}\right|= \)" without writing what it was equal to! Is your problem that you cannot calculate that determinate/cross product?

"Expanding" the determinant by the top row, \(\displaystyle \vec{i}\left|\begin{array}{cc}-15 & -19 \\ -13 & 23\end{array}\right|- \vec{j}\left|\begin{array}{cc} 8 & -19 \\ 2 & 23\end{array}\right|+ \vec{k}\left|\begin{array}{cc}8 & -15 \\ 2 & -13\end{array}\right|= -98\vec{i}- 222\vec{j}- 74\vec{k}\)

In any case, while using the cross product to get a normal vector to the plane is one way to get the equation of the plane, it is not the only way! You should know that any plane, in an xyz-coordinate system, can be written as ax+ by+ cz= d, and that we could always divide through by, say, d, to get ax+ by+ cz= 1 so we need to determine the values of the three coefficients, a, b, and c. We need three equations to do that and we can get three equations by putting the x, y, z coordinates of the three given points in that general equation.

One point is (−3,8,1) so we must have -3a+ 8b+ c= 1.
Another point is (5,−7,−18) so we must have 5a- 7b- 18c= 1.
A third point is (−1,−5,24) so we must have -a- 5b+ 24c= 1.

Solve those three equations for a, b, and c.

 

[karush]

View attachment 7623
this is the example I was trying to follow