[ a b a b ]+ ( a . b )^2= 1. | a |^2 | b |^2 2. | a + b |^2 3. | a |^2+ | b |^2 4. 2 | a |^2 | b |^2

1. | a |^2 | b |^2 Solution: We have [ a b a b ]+ ( a . b )^2 = ( a b ). ( a b )+ ( a . b )^2 = | ( a b ) |^2+ ( a . b )^2 = ( | a | | b | )^2+ ( | a | | b | )^2 = | a |^2 | b |^2 ^2 + | a |^2 | b |^2 ^2 = | a |^2 | b |^2 ( ^2 + ^2 ) = | a |^2 | b |^2

[ a b a b ]+ ( a . b )^2= 1. | a |^2 | b |^2 2. | a + b |^2 3.

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Question

$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=$

$\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2$
$\big|\vec{\text{a}}\big|^2+\big|\vec{\text{b}}\big|^2$
$2\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$

✓

Answer

$\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$

Solution:

We have

$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$

$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$

$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$

$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\theta\big)^2+\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta\big)^2$

$=\big|\vec{\text{a}}\big|^2\big|{\text{b}}\big|^2\sin^2\theta+\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\cos^2\theta$

$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\big(\sin^2\theta+\cos^2\theta\big)$

$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$

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