Let |vec{A}_1| = 3,|vec{A}_2| = 5,and |vec{A} | vedclass

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(C) Given: $|\vec{A}_1| = 3$,$|\vec{A}_2| = 5$,and $|\vec{A}_1 + \vec{A}_2| = 5$.Using the property $|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2

Vedclass · Accepted Answer

$-118$

Vedclass · Answer

(C) Given: $|\vec{A}_1| = 3$,$|\vec{A}_2| = 5$,and $|\vec{A}_1 + \vec{A}_2| = 5$.Using the property $|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta$:$5^2 = 3^2 + 5^2 + 2(3)(5) \cos 	heta$$25 = 9 + 25 + 30 \cos 	heta$$30 \cos 	heta = -9 \implies \vec{A}_1 \cdot \vec{A}_2 = |\vec{A}_1||\vec{A}_2| \cos 	heta = 3 	imes 5 	imes (-9/30) = -4.5$.Now,calculate the dot product:$(2\vec{A}_1 + 3\vec{A}_2) \cdot (3\vec{A}_1 - 2\vec{A}_2) = 6|\vec{A}_1|^2 - 4\vec{A}_1 \cdot \vec{A}_2 + 9\vec{A}_1 \cdot \vec{A}_2 - 6|\vec{A}_2|^2$$= 6|\vec{A}_1|^2 + 5(\vec{A}_1 \cdot \vec{A}_2) - 6|\vec{A}_2|^2$$= 6(3^2) + 5(-4.5) - 6(5^2)$$= 6(9) - 22.5 - 6(25)$$= 54 - 22.5 - 150 = -118.5$.

Column-$I$	Column-$II$
$(a) \|\vec{A} \times \vec{B}\| = 0$	$(i) \theta = 30^{\circ}$
$(b) \|\vec{A} \times \vec{B}\| = 8$	$(ii) \theta = 45^{\circ}$
$(c) \|\vec{A} \times \vec{B}\| = 4$	$(iii) \theta = 90^{\circ}$
$(d) \|\vec{A} \times \vec{B}\| = 4\sqrt{2}$	$(iv) \theta = 0^{\circ}$

Let $|\vec{A}_1| = 3$,$|\vec{A}_2| = 5$,and $|\vec{A}_1 + \vec{A}_2| = 5$. The value of $(2\vec{A}_1 + 3\vec{A}_2) \cdot (3\vec{A}_1 - 2\vec{A}_2)$ is (in $.5$)

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