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 Given: A_1 = 2, A_2 = 3, |A⃗_⃗1⃗ + A⃗_⃗2⃗ | = 3 ⇒ | A⃗_⃗1⃗ + A⃗_⃗2⃗ |^2 = 9 ⇒ A_1^2 + A_2^2 + 2A⃗_⃗1 ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

Given | A⃗⃗...

Question

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

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Solution

The correct option is A $- 64$
Given:
$A_{1} = 2, A_{2} = 3, |_{1} +_{2} | = 3$
$\Rightarrow |_{1} +_{2} |^{2} = 9$
$\Rightarrow A_{1}^{2} + A_{2}^{2} + 2_{1} ._{2} = 9$
On putting value
$\Rightarrow 2^{2} + 3^{2} + 2_{1} ._{2} = 9$
$\Rightarrow_{1} ._{2} = - 2$
Now,
$(_{1} + 2_{2}) . (3_{1} - 4_{2}) = 3 A_{1}^{2} - 8 A_{2}^{2} + 2_{1} ._{2}$
$(_{1} + 2_{2}) . (3_{1} - 4_{2}) = 3 (2)^{2} - 8 (3)^{2} + 2 (- 2) = - 64$

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

The correct option is A −64Given:A1=2,A2=3,|→A1+→A2|=3⇒|→A1+→A2|2=9⇒A21+A22+2→A1.→A2=9On putting value⇒22+32+2→A1.→A2=9⇒→A1.→A2=−2Now, (→A1+2→A2).(3→A1−4→A2)=3A21−8A22+2→A1.→A2(→A1+2→A2).(3→A1−4→A2)=3(2)2−8(3)2+2(−2)=−64

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