Question

If |

⃗A⃗ × B⃗ | = |

⃗A⃗ ∙ B⃗ | then angle between ⃗A⃗ and B⃗ is ……​

Answer 1

Answer:

theta =π/4 or 45°

Explanation:

If we consider θ to be the angle between the vectors A and B,

by defination of cross pr

Therfore, the angle between A and B is π/4 radians.

Answer 2

SOLUTION

GIVEN

$\sf{ | \: \vec{A} \times \vec{B} \: | = | \: \vec{A} \: . \: \vec{B}\: | }$

TO DETERMINE

$\sf{The \: angle \: between \: \: \vec{A} \: \: and \: \: \vec{B} }$

FORMULA TO BE IMPLEMENTED

$\sf{For \: two \: vectors \: \vec{ A} \: \: and \: \vec{B}}$

$\sf{Let \: \: \theta \: \: be \: the \: angle \: between \: \: \vec{A} \: \: and \: \: \vec{B} }$

CROSS PRODUCT

$\sf{ \vec{A} \times \vec{B} = |\vec{A}| \: |\vec{B} | \sin \theta \: \hat{n}}$

Where $\sf{ \hat{n}}$ is a unit vector indicating direction

So that

$\sf{ |\vec{A} \times \vec{B}| = |\vec{A}| \: |\vec{B} | \sin \theta \:}$

DOT PRODUCT

$\sf{ \vec{A} . \vec{B} = |\vec{A}| \: |\vec{B} | \cos \theta }$

So that

$\sf{ | \vec{A} . \vec{B}\: | = |\vec{A}| \: |\vec{B} | \cos \theta }$

EVALUATION

Here it is given that

$\sf{ | \: \vec{A} \times \vec{B} \: | = | \: \vec{A} \: . \: \vec{B}\: | }$

$\sf{ \implies \: \: |\vec{A}| \: |\vec{B} | \sin \theta = | \vec{A}| \: |\vec{B} | \cos \theta\:}$

$\sf{ \implies \: \: \sin \theta = \cos \theta\:}$

$\sf{ \implies \: \: \tan \theta = 1\:}$

$\displaystyle \sf{ \implies \: \theta = \frac{\pi}{4} \:}$

$\displaystyle \: \sf{ \therefore \: \: The \: angle \: between \: \: \vec{A} \: \: and \: \: \vec{B} \: \: is \: \: \frac{\pi}{4} }$

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