MHB Find Vector Perpendicular to Plane

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To find a vector perpendicular to the plane defined by points P (1, 2, 3), Q (2, 3, 1), and R (3, 1, 2), the vector product of vectors PQ and PR is used. The vectors are calculated as PQ = (1, 1, -2) and PR = (2, -1, -1). The cross product PQ × PR is computed using the determinant of a matrix formed by these vectors. This method effectively yields a vector that is orthogonal to the plane formed by the three points. Understanding the vector product is essential for solving this problem.
brinlin
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Find a vector that is perpendicular to the plane passing through the points P (1, 2, 3), Q (2, 3, 1), and R (3, 1, 2).
 
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vector product yields a vector perpendicular to two vectors ...

$\vec{PQ} \times \vec{PR}$
 
I'm sorry I don't really understand.
 
brinlin said:
I'm sorry I don't really understand.

you don't understand, or you don't know what a vector product is and how to calculate it?

$\vec{PQ} = (1,1,-2)$
$\vec{PR} = (2,-1,-1)$

$ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1& 1 & -2 \\ 2 &-1 &-1 \\ \end{vmatrix} $
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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