Cross Product — Question 9

Question 9

Find a vector of magnitude $10$ perpendicular to the plane spanned by $\vec a=\langle1,0,1\rangle$ and $\vec b=\langle0,2,1\rangle$, with positive $z$-component.

Question 9 – 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.

A normal direction is $\vec a\times\vec b=\langle-2,-1,2\rangle$.

Its magnitude is $3$, and its $z$-component is positive.

Scale it to length $10$: $\boxed{\vec n=\frac{10}{3}\langle-2,-1,2\rangle}$.

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.