Question

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)​

Answer 1

Explanation:

The angle is going to 2 times more than actual angle

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

Answer 2

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

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 :-

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

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

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

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