Question

If the position vectors of the vertices \( A, B, C \) of a triangle \( \Delta ABC \) are \( \vec{a}, \vec{b}, \vec{c} \), respectively, then prove that the area of \( \Delta ABC \) is \[ \frac{1}{2} \left| \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \right|. \]

Hint:

Use vector cross products to compute areas of triangles in three-dimensional space.

Answer 1

Theareaofthetriangleisgivenby: \[ \text{Area}=\frac{1}{2}\left|\vec{AB}\times\vec{AC}\right|, \] where\(\vec{AB}=\vec{b}-\vec{a},\vec{AC}=\vec{c}-\vec{a}\).Expandingthecrossproductandsimplifyingyieldstherequiredexpression.