Cross Product — Question 8

Question 8

Prove the area identity $\|\vec u\times\vec v\|^2=\|\vec u\|^2\|\vec v\|^2-(\vec u\cdot\vec v)^2$ using the angle between the vectors.

Question 8 – 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.

If either vector is zero, both sides are zero. Otherwise, let $\theta$ be the angle between $\vec u$ and $\vec v$. Then $\|\vec u\times\vec v\|=\|\vec u\|\|\vec v\|\sin\theta$.

Squaring and using $\sin^2\theta=1-\cos^2\theta$ gives $\|\vec u\|^2\|\vec v\|^2(1-\cos^2\theta)$.

Since $\vec u\cdot\vec v=\|\vec u\|\|\vec v\|\cos\theta$, substitution yields the claimed identity. $\boxed{\text{Proved}}$

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.