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(C) The area of a triangle with vertices having position vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the formula: Area $= \frac{1}{2} |\vec{AB}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The area of a triangle with vertices having position vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the formula: Area $= \frac{1}{2} |\vec{AB} 	imes \vec{AC}|$. Here,$\vec{AB} = \vec{b} - \vec{a}$ and $\vec{AC} = \vec{c} - \vec{a}$. Substituting these into the formula: Area $= \frac{1}{2} |(\vec{b} - \vec{a}) 	imes (\vec{c} - \vec{a})|$. Expanding the cross product: $(\vec{b} - \vec{a}) 	imes (\vec{c} - \vec{a}) = \vec{b} 	imes \vec{c} - \vec{b} 	imes \vec{a} - \vec{a} 	imes \vec{c} + \vec{a} 	imes \vec{a}$. Since $\vec{a} 	imes \vec{a} = 0$,$-\vec{b} 	imes \vec{a} = \vec{a} 	imes \vec{b}$,and $-\vec{a} 	imes \vec{c} = \vec{c} 	imes \vec{a}$,we get: Area $= \frac{1}{2} |\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a}|$. Thus,the correct option is $C$.

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices $A, B, C$ of a triangle respectively. Find the area of the triangle.

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