I have 3 points:
P(-3, 1, 2), Q(-1, 2, 3), R(2, 1, 0)
and I need to find a nonzero vector orthogonal to the plane through these three points.
I seem to recall this having something to do with the cross product, so I mad vectors PQ <2,1,1> & PR <5,0,-2> calculated PQxPR and got the vector <0,9,-5>.
What do I do now?
Thank you in advance.
Your cross product vector (for PQ x PR), if done correctly, is a valid answer.
Any vector proportional to <0,9,-5> is aso a valid answer.
You're correct! To find a nonzero vector orthogonal to the plane through three points, you can use the cross product.
In your case, you correctly calculated the vector PQ as <2, 1, 1> and PR as <5, 0, -2>. To find a vector orthogonal to the plane containing these points, you can take the cross product of PQ and PR.
The cross product can be calculated by finding the determinant of a 3x3 matrix. Here's how you can do it:
1. Write out the two vectors, PQ and PR, as a 3x3 matrix:
| i j k |
| 2 1 1 |
| 5 0 -2 |
2. Calculate the determinant of this matrix (also known as the cross product):
(1 * -2 - 1 * 0)i - (2 * -2 - 1 * 5)j + (2 * 0 - 1 * 5)k
= -2i - (-4j) + (-5k)
= -2i + 4j - 5k
So the cross product of PQ and PR is the vector <-2, 4, -5>. This vector is orthogonal (perpendicular) to the plane containing the three points P, Q, and R.
Note that the vector you calculated, <0, 9, -5>, is scalar multiple of the correct orthogonal vector. You can confirm this by dividing the vector components by a non-zero scalar. In your case, dividing by 9 gives the correct orthogonal vector <-2/9, 4/9, -5/9>.
I hope this clarifies the next steps for you! Let me know if you have any more questions.