# 243.12.7.12 plane from 3 points

• MHB
• karush
In summary, the equation for the plane through points is: $\vec{AB}\times \vec{AC}= \begin{bmatrix} \textbf{i} & \textbf{j} & \textbf{k}\\\, 5 &-7 &-18\\-1 &-5 &24 \end{bmatrix}$
karush
Gold Member
MHB
$\tiny\textit{243.12.7.12}$

$\textsf{Write the equation for the plane through points}$
$$A(-3,8,1),\, B(5,-7,-18),\, C(-1,-5,24)$$
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 $\color{red}{ \displaystyle 8x + 3y + z = 1 }$

$$ax+by+cz=d$$

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

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 $C$.

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

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.

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## What is the definition of a "243.12.7.12 plane from 3 points"?

A "243.12.7.12 plane from 3 points" refers to a mathematical concept in three-dimensional space where a plane is defined by three distinct points that lie on the same plane.

## How is a "243.12.7.12 plane from 3 points" represented mathematically?

In mathematical notation, a "243.12.7.12 plane from 3 points" can be represented by the equation ax + by + cz + d = 0, where a, b, and c are the coordinates of the three points and d is a constant.

## What is the significance of the numbers 243.12.7.12 in a "243.12.7.12 plane from 3 points"?

The numbers 243.12.7.12 represent the coordinates of the three points used to define the plane. Each number corresponds to a specific axis in three-dimensional space (x, y, and z) and determines the position of the point along that axis.

## How is a "243.12.7.12 plane from 3 points" different from a plane defined by two points and a vector?

A "243.12.7.12 plane from 3 points" is defined by the actual points on the plane, whereas a plane defined by two points and a vector is defined by the two points and the direction of the plane, which is determined by the vector.

## Can a "243.12.7.12 plane from 3 points" exist in any orientation in three-dimensional space?

Yes, a "243.12.7.12 plane from 3 points" can exist in any orientation in three-dimensional space as long as the three points are not collinear (lie on the same line).

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