Cross Product — Question 3

Question 3 – Solution

Keep the component order and signs organized when expanding the determinant. After computing the cross product, use a dot-product check or the relevant magnitude formula to interpret it.

Take a cross product: $\langle2,1,0\rangle\times\langle1,-1,3\rangle=\langle3,-6,-3\rangle=3\langle1,-2,-1\rangle$.

Its magnitude is $3\sqrt6$.

Normalize and include both orientations: $\boxed{\pm\frac1{\sqrt6}\langle1,-2,-1\rangle}$.

The result follows from the defining vector formulas used above, and each component, magnitude, or scalar condition has been checked against the information in the question.

The cross product produces a vector perpendicular to both inputs. Its direction follows the right-hand rule, while its magnitude records the area of the spanned parallelogram.