Cross Product — Question 2

Question 2

Find the area of the triangle with vertices $A=(1,0,2)$, $B=(3,-1,4)$, and $C=(0,2,5)$.

Question 2 – 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.

Form two sides: $\overrightarrow{AB}=\langle2,-1,2\rangle$ and $\overrightarrow{AC}=\langle-1,2,3\rangle$.

Their cross product is $\langle-7,-8,3\rangle$, whose magnitude is $\sqrt{49+64+9}=\sqrt{122}$.

A triangle has half the parallelogram area, so $\boxed{A=\sqrt{122}/2}$.

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.