If |bar{a}|=2, |bar{b}|=3 and bar{a}, bar{b} | vedclass

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(B) Let the vertices of the triangle be $O(0), A(\bar{a}+2\bar{b}), B(\bar{a}-2\bar{b})$.The area of the triangle is given by $\frac{1}{2} |\vec{OA} \

Vedclass · Accepted Answer

$12 	ext{ sq. units}$

Vedclass · Answer

(B) Let the vertices of the triangle be $O(0), A(\bar{a}+2\bar{b}), B(\bar{a}-2\bar{b})$.The area of the triangle is given by $\frac{1}{2} |\vec{OA} 	imes \vec{OB}|$.$\vec{OA} = \bar{a}+2\bar{b}$ and $\vec{OB} = \bar{a}-2\bar{b}$.$\vec{OA} 	imes \vec{OB} = (\bar{a}+2\bar{b}) 	imes (\bar{a}-2\bar{b})$$= \bar{a} 	imes \bar{a} - 2(\bar{a} 	imes \bar{b}) + 2(\bar{b} 	imes \bar{a}) - 4(\bar{b} 	imes \bar{b})$Since $\bar{a} 	imes \bar{a} = 0$ and $\bar{b} 	imes \bar{b} = 0$,and $\bar{b} 	imes \bar{a} = -(\bar{a} 	imes \bar{b})$:$= 0 - 2(\bar{a} 	imes \bar{b}) - 2(\bar{a} 	imes \bar{b}) - 0 = -4(\bar{a} 	imes \bar{b})$.Area $= \frac{1}{2} |-4(\bar{a} 	imes \bar{b})| = 2 |\bar{a} 	imes \bar{b}|$.Since $\bar{a} \perp \bar{b}$,$|\bar{a} 	imes \bar{b}| = |\bar{a}| |\bar{b}| \sin(90^{\circ}) = 2 	imes 3 	imes 1 = 6$.Area $= 2 	imes 6 = 12 	ext{ sq. units}$.

If $|\bar{a}|=2, |\bar{b}|=3$ and $\bar{a}, \bar{b}$ are mutually perpendicular vectors,then the area of the triangle whose vertices are $0, \bar{a}+2\bar{b}, \bar{a}-2\bar{b}$ is

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