Physics, asked by diya897, 2 months ago

36. A and B are two non-zero vectors.
a) If AXB = Ā.B , what is the angle between A and B ?
b) Find the value of Ā .(A xB)​

Answers

Answered by raorita1981
0

Explanation:

The angle is going to 2 times more than actual angle

value of b is always sum of 1/3 of of product

Answered by mathdude500
13

\large\underline{\sf{Solution-(a)}}

We know,

 \boxed{ \sf \:  |\vec{a} \times \vec{b}|=  | \vec{a}| | \vec{b}|sin \theta}

and

 \boxed{ \sf \: \vec{a} \: . \: \vec{b} =  |\vec{a}| \: |\vec{b}| \: cos\theta}

According to statement,

\rm :\longmapsto\: |\vec{a} \times \vec{b}|  = \vec{a} \: . \: \vec{b}

\rm :\longmapsto\: |\vec{a}| |\vec{b}|sin\theta =  |\vec{a}| |\vec{b}|cos\theta

\rm :\longmapsto\:sin\theta = cos\theta

\rm :\longmapsto\:tan\theta = 1

\bf\implies \:\theta = \dfrac{\pi}{4}

\large\underline{\sf{Solution-(b)}}

We know that

 \boxed{ \sf \: [\vec{a} \: \vec{b} \: \vec{c}] = \vec{a}. \: (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}). \: \vec{c}}

 \sf \:  \therefore \: \vec{a} \: . \: (\vec{a} \times \vec{b}) = (\vec{a} \times \vec{a}) \: . \: \vec{b} =\vec{0}  \: . \: \vec{b} =0

\rm :\implies\:\vec{a}.(\vec{a} \times \vec{b})\: is \: coplanar.

\bf\implies \:Angle \: between \: \vec{a}.(\vec{a} \times \vec{b}) = 0 \degree

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Additional Information :-

 \boxed{ \sf \: \vec{a} \: . \: \vec{a} =  { |\vec{a}| }^{2}}

 \boxed{ \sf \: \vec{a} \: . \: \vec{b} \:  =  \: \vec{b} \: . \: \vec{a}}

 \boxed{ \sf \: \vec{a} \:  \times  \: \vec{a} \:  =  \: 0}

 \boxed{ \sf \: \vec{a} \times \vec{b} =  -  \: \vec{b} \times \vec{a}}

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