Let $|\vec{A_{1}}| = 3 |\vec{A_{2}}| = 5$ and $|\vec{A_{1}} + \vec{A_{2}}| = 5 .$ The value of $(2 \vec{A_{1}} + 3 \vec{A_{2}}) . (3 \vec{A_{1}} - 2 \vec{A_{2}})$ is.

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–106.5

–112.5

–118.5

–99.5

answer is D.

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Detailed Solution

Given,

$|\vec{A_{1}}| = 3, |\vec{A_{2}}| = 5 & |\vec{A_{1}} + \vec{A_{2}}| = 5 .$

We know that,

$|\vec{A_{1}} + \vec{A_{2}}| = \sqrt{|\vec{A_{1}} | + | \vec{A_{2}}| + 2 |\vec{A_{1}} | | \vec{A_{2}}| \cos θ}$

$\Rightarrow 5 = \sqrt{9 + 25 + 2 \times 3 \times 5 \cos θ}$

$\Rightarrow \cos θ = - \frac{9}{2 \times 3 \times 5} = - \frac{3}{10}$

Now,

$(2 \vec{A_{1}} + 3 \vec{A_{2}}) . (3 \vec{A_{1}} - 2 \vec{A_{2}}) = 6 | \vec{A_{1}} |^{2} + 9 \vec{A_{1}} . \vec{A_{2}} - 6 | \vec{A_{2}} |^{2}$

$= 54 + 5 \times 3 \times (\frac{- 3}{10}) - 6 \times 25$

$= 54 - 150 - \frac{45}{2} = - 118.5$

Hence the correct answer is -118.5.

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