Given |vec{A}_1|=2, |vec{A}_2|=3 and |vec{A}_ | vedclass

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(A) Given that $|\vec{A}_1|=2, |\vec{A}_2|=3$ and $|\vec{A}_1+\vec{A}_2|=3$. Squaring the magnitude of the sum: $(|\vec{A}_1+\vec{A}_2|)^2 = 3^2$ $A_1

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$-64$

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(A) Given that $|\vec{A}_1|=2, |\vec{A}_2|=3$ and $|\vec{A}_1+\vec{A}_2|=3$. Squaring the magnitude of the sum: $(|\vec{A}_1+\vec{A}_2|)^2 = 3^2$ $A_1^2 + A_2^2 + 2 \vec{A}_1 \cdot \vec{A}_2 = 9$ $4 + 9 + 2 \vec{A}_1 \cdot \vec{A}_2 = 9$ $13 + 2 \vec{A}_1 \cdot \vec{A}_2 = 9$ $2 \vec{A}_1 \cdot \vec{A}_2 = -4$ $\vec{A}_1 \cdot \vec{A}_2 = -2$ Now,expand the expression $(\vec{A}_1+2 \vec{A}_2) \cdot (3 \vec{A}_1-4 \vec{A}_2)$: $= 3 \vec{A}_1 \cdot \vec{A}_1 - 4 \vec{A}_1 \cdot \vec{A}_2 + 6 \vec{A}_2 \cdot \vec{A}_1 - 8 \vec{A}_2 \cdot \vec{A}_2$ $= 3 |\vec{A}_1|^2 + 2 \vec{A}_1 \cdot \vec{A}_2 - 8 |\vec{A}_2|^2$ Substitute the values: $= 3(2)^2 + 2(-2) - 8(3)^2$ $= 3(4) - 4 - 8(9)$ $= 12 - 4 - 72$ $= 8 - 72 = -64$

Given $|\vec{A}_1|=2, |\vec{A}_2|=3$ and $|\vec{A}_1+\vec{A}_2|=3$. Find the value of $(\vec{A}_1+2 \vec{A}_2) \cdot (3 \vec{A}_1-4 \vec{A}_2)$ :

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