Cross Product — Question 1

Question 1 – 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.

Expand the determinant: $\vec a\times\vec b=\langle2(4)-(-1)(0),\,(-1)(3)-1(4),\,1(0)-2(3)\rangle=\langle8,-7,-6\rangle$.

Check $(\vec a\times\vec b)\cdot\vec a=8-14+6=0$.

Also $(\vec a\times\vec b)\cdot\vec b=24-24=0$, so $\boxed{\vec a\times\vec b=\langle8,-7,-6\rangle}$ is perpendicular to both.

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.