Given -A-1- - 2- -A-2- 3 and -A-1 - A-2- - 3- Find the value of A-1 - 2A-2-3A-1 -4A-2-

The correct option is A 64 | A⃗ #160; _ 1 | =2 | A⃗ #160; _ 2 | =3 | A⃗ #160; _ 1 +A⃗ #160; _ 2 | =3 ( A⃗ #160; _ 1 +2A⃗ #160; _ ...

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Byju's Answer

Standard XII

Mathematics

Addition of Vectors

Given |A⃗1|...

Question

Given $| {\to A}_{1} | = 2$ , $| {\to A}_{2} | = 3$ and $| {\to A}_{1} + {\to A}_{2} | = 3$ . Find the value of $({\to A}_{1} + 2 {\to A}_{2}) \cdot (3 {\to A}_{1} - 4 {\to A}_{2})$ .

- 64

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Solution

The correct option is A $- 64$
$∣ ∣ {\to A}_{1} ∣ ∣ = 2 ∣ ∣ {\to A}_{2} ∣ ∣ = 3 ∣ ∣ {\to A}_{1} + {\to A}_{2} ∣ ∣ = 3$
$({\to A}_{1} + 2 {\to A}_{2}) . (3 {\to A}_{1} - 4 {\to A}_{2}) = ?$
$3 = \sqrt{4 + 9 + 2 \times 6 cos θ}$
$9 = 4 + 9 + 12 cos θ \Rightarrow \frac{- 4}{12} = cos θ$
$\Rightarrow c o s θ = - 1$
$({\to A}_{1} .2 {\to A}_{2}) . (3 {\to A}_{1} - 4^{3} {\to A}_{2})$
$= 3 {| A_{1} |}^{2} - 4 {\to A}_{1} . {\to A}_{2} + 6. {\to A}_{1} . {\to A}_{2} - 8 {∣ ∣ {\to A}_{2} ∣ ∣}^{2}$
$= 3 {(2)}^{2} - 4 \times 2 \times 3 cos θ + 6 \times 2 \times 3 cos θ - 8 \times {(3)}^{2}$
$= 12 + 8 - 12 - 8 \times 9$
$= - 72 + 8$
$= - 64$

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Given |→A1|=2, |→A2|=3 and |→A1+→A2|=3. Find the value of (→A1+2→A2)⋅(3→A1−4→A2).

Given $| {\to A}_{1} | = 2$ , $| {\to A}_{2} | = 3$ and $| {\to A}_{1} + {\to A}_{2} | = 3$ . Find the value of $({\to A}_{1} + 2 {\to A}_{2}) \cdot (3 {\to A}_{1} - 4 {\to A}_{2})$ .