If |a| = 2,|b| = 3 and a, b are mutually perp | vedclass

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

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Let the vertices of the triangle be $O(0)$,$A(a + b)$,and $B(a - b)$.The area of the triangle is given by $\frac{1}{2} |\vec{OA} 	imes \vec{OB}|$.Substituting the vectors,we get $	ext{Area} = \frac{1}{2} |(a + b) 	imes (a - b)|$.Expanding the cross product: $(a + b) 	imes (a - b) = a 	imes a - a 	imes b + b 	imes a - b 	imes b$.Since $a 	imes a = 0$ and $b 	imes b = 0$,and $b 	imes a = -(a 	imes b)$,we have:$(a + b) 	imes (a - b) = 0 - (a 	imes b) - (a 	imes b) - 0 = -2(a 	imes b) = 2(b 	imes a)$.Thus,$	ext{Area} = \frac{1}{2} |2(b 	imes a)| = |b 	imes a| = |b| |a| \sin(90^\circ)$.Given $|a| = 2$ and $|b| = 3$,and the vectors are perpendicular $(	heta = 90^\circ)$:$	ext{Area} = 3 	imes 2 	imes 1 = 6$.

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

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