Geometric Interpretation of Cross Product

Geometric Interpretation of Cross Product

1. $$\vec{a}\times \vec{b}=|\vec{a}||\vec{b}|\sin \theta \hat{n}$$.

$$|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta$$.

$$=2\left( \frac{1}{2}\times |\vec{a}||\vec{b}|\sin\theta\right)$$.

= 2 (Area of Triangle AOC)

= Area of Parallelogram

Area of the triangle OAB is $$\frac{1}{2}|\vec{a}\times \vec{b}|$$.

$$\vec{a}\times \vec{b}$$ is said to be the vector area of the parallelogram with adjacent sides OA and OB.

2. If $$\vec{a},\vec{b}$$ are diagonals of a parallelogram, its area = $$\frac{1}{2}|\vec{a}\times \vec{b}|$$.

In fig $$\overrightarrow{OC}=\overrightarrow{a}$$ and $$\overrightarrow{AB}=\overrightarrow{b}$$.

Area parallelogram = $$4\times \frac{1}{2}|\overrightarrow{PC}\times \overrightarrow{PB}|$$.

= $$4\times \frac{1}{2}\left| \frac{{\vec{a}}}{2}\times \frac{{\vec{b}}}{2} \right|$$.

$$=\frac{1}{2}\left| \vec{a}\times \vec{b} \right|$$.

3. If AC and BD are the diagonals of a quadrilateral, then its vector area is $$\frac{1}{2}\overrightarrow{AC}\times \overrightarrow{BD}$$.

Vector area of the quadrilateral ABCD = Vector area of ΔABC + Vector area of ΔACD.

$$=\frac{1}{2}\overrightarrow{AB}\times \overrightarrow{AC}+\frac{1}{2}\overrightarrow{AC}\times \overrightarrow{AD}$$.

$$=-\frac{1}{2}\overrightarrow{AC}\times \overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\times \overrightarrow{AD}$$.

$$=\frac{1}{2}\overrightarrow{AC}\times (\overrightarrow{AD}-\overrightarrow{AB})$$.

$$=\frac{1}{2}\overrightarrow{AC}\times \overrightarrow{BD}$$.