A-1 and A-2 are two vectors such that -A-1- - 3- -A-2- - 5 and -A-1 - A-2- - 5 the value of 2A-1 - 3A-2 - 2A-1 - 2A-2 is

The correct option is B 123 (A⃗_1 + A⃗_2) · (A⃗_1 + A⃗_2) = |A⃗_1 + A⃗_2|^2 9 + 25 + 2 (A⃗_1 ·A⃗_2) = 25 (A⃗_1 ·A⃗_2) = 92 (2A⃗_1 + ...

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Byju's Answer

Standard XII

Physics

Vector Component

A⃗1 and A⃗2...

Question

${\to A}_{1}$ and ${\to A}_{2}$ are two vectors such that $| {\to A}_{1} | = 3, | {\to A}_{2} | = 5$ and $| {\to A}_{1} + {\to A}_{2} | = 5$ the value of $(2 {\to A}_{1} + 3 {\to A}_{2}) \cdot (2 {\to A}_{1} - 2 {\to A}_{2})$ is

\frac{237}{2}

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- 123

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\frac{- 337}{2}

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\frac{337}{2}

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Solution

The correct option is B $- 123$
$({\to A}_{1} + {\to A}_{2}) \cdot ({\to A}_{1} + {\to A}_{2}) = | {\to A}_{1} + {\to A}_{2} |^{2}$
$9 + 25 + 2 ({\to A}_{1} \cdot {\to A}_{2}) = 25$
$({\to A}_{1} \cdot {\to A}_{2}) = \frac{- 9}{2}$
$(2 {\to A}_{1} + 3 {\to A}_{2}) \cdot (2 {\to A}_{1} - 2 {\to A}_{2}) = 4 | {\to A}_{1} |^{2} - 4 {\to A}_{1} \cdot {\to A}_{2} + 6 {\to A}_{2} \cdot {\to A}_{1} = 6 | {\to A}_{2} |^{2}$
$= 4 A_{1}^{2} - 4 A_{1} \cdot A_{2} + 6 A_{1} \cdot A_{2} - 6 A_{2}^{2}$
$= 4 \times 9 + 2 \times (\frac{- 9}{2}) - 6 \times 25 = - 123$

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→A1 and →A2 are two vectors such that |→A1|=3,|→A2|=5 and |→A1+→A2|=5 the value of (2→A1+3→A2)⋅(2→A1−2→A2) is

The correct option is B −123(→A1+→A2)⋅(→A1+→A2)=|→A1+→A2|29+25+2(→A1⋅→A2)=25(→A1⋅→A2)=−92(2→A1+3→A2)⋅(2→A1−2→A2)=4|→A1|2−4→A1⋅→A2+6→A2⋅→A1=6|→A2|2 =4A21−4A1⋅A2+6A1⋅A2−6A22 =4×9+2×(−92)−6×25=−123

${\to A}_{1}$ and ${\to A}_{2}$ are two vectors such that $| {\to A}_{1} | = 3, | {\to A}_{2} | = 5$ and $| {\to A}_{1} + {\to A}_{2} | = 5$ the value of $(2 {\to A}_{1} + 3 {\to A}_{2}) \cdot (2 {\to A}_{1} - 2 {\to A}_{2})$ is