How to Find the Equation of a Plane Through Three Points

[whig4life]

containing: (1,1,2) (-1,3,5) (2,-1,4)

Here is what I have done, and where I am stuck:

P= (1,1,2) Q= (-1,3,5) R= (2,-1,4)

PQ= (2,-2,-3) and QR=(-3,4,1)

n= PQ x QR = (a, b, c) = (-14,-11,2) while r0=(1,1,2)


-14x-11y+2z+21=12 for f(Q) and f(R)...f(P) was zero. I am assuming d=12 although I found it difficult to be certain.

So I am wondering if my plane equation is bad or if I got the wrong "d"?

In general, where did I go wrong and what should I remember in the future?

 

[HallsofIvy]

containing: (1,1,2) (-1,3,5) (2,-1,4)

Here is what I have done, and where I am stuck:

P= (1,1,2) Q= (-1,3,5) R= (2,-1,4)

PQ= (2,-2,-3) and QR=(-3,4,1)

n= PQ x QR = (a, b, c) = (-14,-11,2)

This incorrect.
\(\displaystyle PQ x QR= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 2 & - 2 & -3 \\ -3 & 4 & 1 \end{array}\right|= \vec{i}\left|\begin{array}{cc}-2 & -3 \\ 4 & 1 \end{array}\right|- \vec{j}\left|\begin{array}{cc} 2 & -3 \\ -3 & 1\end{array}\right|+ \vec{k}\left|\begin{array}{cc}2 & -2 \\ 4 & 1 \end{array}\right|\)= \(\displaystyle \vec{i}(-2-(12))- \vec{j}(2- 12)+ \vec{k}(8- 6)= 10\vec{i}+ 10\vec{j}+ 2\vec{k}\).
You appear to have missed some negative signs.

f(Q) and f(R)...f(P) was zero. I am assuming d=12 although I found it difficult to be certain.

So I am wondering if my plane equation is bad or if I got the wrong "d"?

In general, where did I go wrong and what should I remember in the future?

So one way of doing this is to write 10x+10y+ 2z= d where you can determine d by putting any one of the given points in for x, y, and z. Equivalently, use the formula \(\displaystyle 10(x- x_0)+ 10(y- y_0)+ 2(z- z_0)= 0\) where \(\displaystyle (x_0, y_0, z_0)\) is any one of the given points. Perhaps you are confusing those two methods.

 

[whig4life]

This incorrect.
\(\displaystyle PQ x QR= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 2 & - 2 & -3 \\ -3 & 4 & 1 \end{array}\right|= \vec{i}\left|\begin{array}{cc}-2 & -3 \\ 4 & 1 \end{array}\right|- \vec{j}\left|\begin{array}{cc} 2 & -3 \\ -3 & 1\end{array}\right|+ \vec{k}\left|\begin{array}{cc}2 & -2 \\ 4 & 1 \end{array}\right|\)= \(\displaystyle \vec{i}(-2-(12))- \vec{j}(2- 12)+ \vec{k}(8- 6)= 10\vec{i}+ 10\vec{j}+ 2\vec{k}\).
You appear to have missed some negative signs.

r f(Q) and f(R)...f(P) was zero. I am assuming d=12 although I found it difficult to be certain.

So I am wondering if my plane equation is bad or if I got the wrong "d"?

In general, where did I go wrong and what should I remember in the future?

So one way of doing this is to write 10x+10y+ 2z= d where you can determine d by putting any one of the given points in for x, y, and z. Equivalently, use the formula \(\displaystyle 10(x- x_0)+ 10(y- y_0)+ 2(z- z_0)= 0\) where \(\displaystyle (x_0, y_0, z_0)\) is any one of the given points. Perhaps you are confusing those two methods.[/QUOTE]

You are right I missed a negative sign, but I am getting (10,7,2) for vector n via PQ xQR.

 

[HallsofIvy]

\(\displaystyle PQ x QR= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 2 & - 2 & -3 \\ -3 & 4 & 1 \end{array}\right|= \vec{i}\left|\begin{array}{cc}-2 & -3 \\ 4 & 1 \end{array}\right|- \vec{j}\left|\begin{array}{cc} 2 & -3 \\ -3 & 1\end{array}\right|+ \vec{k}\left|\begin{array}{cc}2 & -2 \\ 4 & 1 \end{array}\right|\)= \(\displaystyle \vec{i}(-2-(12))- \vec{j}(2- 12)+ \vec{k}(8- 6)= 10\vec{i}+ 10\vec{j}+ 2\vec{k}\).

Blast! That should be \(\displaystyle \vec{i}(-2-(-12))- \vec{j}(2- 9)+ \vec{k}(2+ 8)= 10\vec{i}- 7\vec{j}+ 10\vec{k}\)

You appear to have missed some negative signs.



So one way of doing this is to write 10x+10y+ 2z= d where you can determine d by putting any one of the given points in for x, y, and z. Equivalently, use the formula \(\displaystyle 10(x- x_0)+ 10(y- y_0)+ 2(z- z_0)= 0\) where \(\displaystyle (x_0, y_0, z_0)\) is any one of the given points. Perhaps you are confusing those two methods.