Vector to Scalar

[Alexmcom]

Sure. I don't mind. I might not know the answer, but I'm certainly willing to give it a go.

Awesome, thanks dude!

So a week ago I messaged my teacher asking something simple but I am still left confused. This has to do with creating a scalar equation.
For example; determine the scalar equation of P(1,0,4) Q(3,1,-6) and R(-2,3,5). Usually you have to subtract one another to create points for the cross multiplication right. However doing them in different order yields different answers and my teacher said that it should be okay.

His words ''Yes, absolutely, the order doesn't matter. It yields the same answer regardless. If you were to do AB and AC, it would yield the same answer as AB and BC. Give it a try to see how it works.''

So for example PQ=Q-P=(3,1,6)-(1,0,4)=3-1,1-0,6-4=(2,1,2)
PR=R-P=(-2,3,5)-(1,0,4)=2-1,3-0,5-4=(1,3,1)

Cross multiply (PQxPR)= (1)(1)-(2)(3),(2)(1)-(2)(1),(2)(3)-(1)(1)=1-6,2-2,6-1=-5,0,5

Now if we did this for PQ and QR it would be like this;

PQ=(2,1,2)
QR=R-Q=(-2,3,5)-(3,1,-6)=-2-3,3-1,5+6=-5,2,11

Cross multiply them together (PQxQR)=(2)(2)-(11)(1),(11)(2)-(-5)(2),(-5)(1)-(2)(2)=-7,32,-9



I rushed this so if the answers are incorrect I apologize. So I've noticed that when I Add subtract the points in another order I get another answer.

 

[ksdhart2]

It's been a while since I've done this and I'm a bit rusty, but from what I remember, what your professor said should be true. The order shouldn't matter. Okay, upon a closer examination, I've figured out what's going on. Your error isn't in the cross product, but rather just some sloppiness involving sign errors.

Given the points P(1,0,4), Q(3,1,-6), and R(-2,3,5), \(\displaystyle \overrightarrow{PQ}=<3-1,1-0,\boldsymbol{\color{red}{-}}6-4>=<2,1,-10>\), \(\displaystyle \overrightarrow{PR}=<\boldsymbol{\color{red}{-}}2,3-0,5-4>=<-3,3,1>\), and \(\displaystyle \overrightarrow{QR}=<-2-3,3-1,5-(-6)>=<-5,2,11>\)

Then if we take the cross product, either of PQ and PR or PQ and QR, we should get the same answer (or the same values but with a sign difference). But let's double check:

\(\displaystyle \overrightarrow{PQ} \times \overrightarrow{PR}=\text{det} \begin{pmatrix}\vec{i}&\vec{j}&\vec{k}\\ 2&1&-10\\ -3&3&1\end{pmatrix}=<-31,-28,-9>\)

\(\displaystyle \overrightarrow{PQ} \times \overrightarrow{QR}=\text{det} \begin{pmatrix}\vec{i}&\vec{j}&\vec{k}\\ 2&1&-10\\ -5&2&11\end{pmatrix}=<31,28,9>\)

This checks out. \(\displaystyle \overrightarrow{PQ} \times \overrightarrow{PR}=-\left(\overrightarrow{PQ} \times \overrightarrow{QR}\right)\). Either one of these vectors will be normal (one's oriented up and the other down) to the plane, and we can use any of the three given points to find the equation of the plane.

 

[Alexmcom]

It's been a while since I've done this and I'm a bit rusty, but from what I remember, what your professor said should be true. The order shouldn't matter. Okay, upon a closer examination, I've figured out what's going on. Your error isn't in the cross product, but rather just some sloppiness involving sign errors.

Given the points P(1,0,4), Q(3,1,-6), and R(-2,3,5), \(\displaystyle \overrightarrow{PQ}=<3-1,1-0,\boldsymbol{\color{red}{-}}6-4>=<2,1,-10>\), \(\displaystyle \overrightarrow{PR}=<\boldsymbol{\color{red}{-}}2,3-0,5-4>=<-3,3,1>\), and \(\displaystyle \overrightarrow{QR}=<-2-3,3-1,5-(-6)>=<-5,2,11>\)

Then if we take the cross product, either of PQ and PR or PQ and QR, we should get the same answer (or the same values but with a sign difference). But let's double check:

\(\displaystyle \overrightarrow{PQ} \times \overrightarrow{PR}=\text{det} \begin{pmatrix}\vec{i}&\vec{j}&\vec{k}\\ 2&1&-10\\ -3&3&1\end{pmatrix}=<-31,-28,-9>\)

\(\displaystyle \overrightarrow{PQ} \times \overrightarrow{QR}=\text{det} \begin{pmatrix}\vec{i}&\vec{j}&\vec{k}\\ 2&1&-10\\ -5&2&11\end{pmatrix}=<31,28,9>\)

This checks out. \(\displaystyle \overrightarrow{PQ} \times \overrightarrow{PR}=-\left(\overrightarrow{PQ} \times \overrightarrow{QR}\right)\). Either one of these vectors will be normal (one's oriented up and the other down) to the plane, and we can use any of the three given points to find the equation of the plane.

Dude thank you, I appreciate it! It makes a lot more sense. I should better use the calculator all the time and re-read the question more carefully for this just to be safe

 

[ksdhart2]

Yeah, I know what you mean. It used to be, back in the day, I was so certain, but now I find myself using a calculator to find 3 times 5, "just to be sure." For me, and a lot of my classmates it seems, it's often the little stuff that trips us up, like missing a minus sign, or somehow thinking 2 times 3 is 5.