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(B) The vector area of a triangle with vertices having position vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ is given by the f

Vedclass · Accepted Answer

$\frac{1}{2}\{\overrightarrow{a} 	imes \overrightarrow{b}+\overrightarrow{b} 	imes \overrightarrow{c}+\overrightarrow{c} 	imes \overrightarrow{a}\}$

Vedclass · Answer

(B) The vector area of a triangle with vertices having position vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ is given by the formula: $	ext{Vector Area} = \frac{1}{2} (\overrightarrow{AB} 	imes \overrightarrow{AC})$ Since $\overrightarrow{AB} = \overrightarrow{b} - \overrightarrow{a}$ and $\overrightarrow{AC} = \overrightarrow{c} - \overrightarrow{a}$, we have: $	ext{Vector Area} = \frac{1}{2} ((\overrightarrow{b} - \overrightarrow{a}) 	imes (\overrightarrow{c} - \overrightarrow{a}))$ Expanding the cross product: $	ext{Vector Area} = \frac{1}{2} (\overrightarrow{b} 	imes \overrightarrow{c} - \overrightarrow{b} 	imes \overrightarrow{a} - \overrightarrow{a} 	imes \overrightarrow{c} + \overrightarrow{a} 	imes \overrightarrow{a})$ Since $\overrightarrow{a} 	imes \overrightarrow{a} = 0$, $-\overrightarrow{b} 	imes \overrightarrow{a} = \overrightarrow{a} 	imes \overrightarrow{b}$, and $-\overrightarrow{a} 	imes \overrightarrow{c} = \overrightarrow{c} 	imes \overrightarrow{a}$, we get: $	ext{Vector Area} = \frac{1}{2} (\overrightarrow{a} 	imes \overrightarrow{b} + \overrightarrow{b} 	imes \overrightarrow{c} + \overrightarrow{c} 	imes \overrightarrow{a})$

Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be the position vectors of the vertices $A, B, C$ respectively of $\triangle ABC$. The vector area of $\triangle ABC$ is:

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