The value of ( a b )^2 is: 1. | a |^2+ | b |^2- ( a . b )^2 2. | a |^2 | b |^2- ( a . b )^2 3. | a |^2+ | b |^2-2 ( a . b ) 4. | a |^2+ | b |^2- a . b

2. | a |^2 | b |^2- ( a . b )^2 Solution: ( a . b )^2+ | a b |^2 = (| a | | b | )^2+ (| a | | b | )^2 =| a |^2 | b |^2( ^2 + ^2 ) =| a |^2 | b |^2 (1) =| a |^2 | b |^2 | a b |^2=| a |^2 | b |^2- ( a . b )^2 Thus, the value of ( a b )^2 is | a |^2 | b |^2- ( a . b )^2.

The value of ( a b )^2 is: 1. | a |^2+ | b |^2- ( a . b )^2 2.

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Question

The value of $\big(\vec{\text{a}}\times\vec{\text{b}}\big)^2$ is:

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

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Answer

$|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$

Solution:

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

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

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

$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$ (1)

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

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

Thus, the value of $\big(\vec{\text{a}}\times\vec{\text{b}}\big)^2$ is $|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2.$

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VECTOR ALGEBRA — Maths STD 12 Science — Question

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