Answer :
To find the angle between vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] given the relation [tex]\(\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|\)[/tex], we can use the properties of the dot product and cross product. Let's break it down step-by-step:
1. Dot Product:
The dot product of two vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] is given by:
[tex]\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \][/tex]
where [tex]\(\theta\)[/tex] is the angle between the vectors, and [tex]\(|\vec{A}|\)[/tex] and [tex]\(|\vec{B}|\)[/tex] are the magnitudes of [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] respectively.
2. Cross Product:
The magnitude of the cross product of [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] is given by:
[tex]\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \][/tex]
3. Given Condition:
According to the problem, we have:
[tex]\[ \vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}| \][/tex]
4. Substitute the Dot Product and Cross Product Formulas:
Substitute the expressions for the dot product and magnitude of the cross product:
[tex]\[ |\vec{A}| |\vec{B}| \cos \theta = |\vec{A}| |\vec{B}| \sin \theta \][/tex]
5. Simplify the Equation:
If both [tex]\(|\vec{A}|\)[/tex] and [tex]\(|\vec{B}|\)[/tex] are non-zero, we can divide both sides of the equation by [tex]\(|\vec{A}| |\vec{B}|\)[/tex]:
[tex]\[ \cos \theta = \sin \theta \][/tex]
6. Solve for [tex]\(\theta\)[/tex]:
Recall the trigonometric identity that [tex]\(\cos \theta = \sin \theta\)[/tex] when [tex]\(\theta\)[/tex] is 45 degrees. Therefore:
[tex]\[ \tan \theta = 1 \][/tex]
The angle whose tangent is 1 is:
[tex]\[ \theta = 45^\circ \][/tex]
Therefore, the angle between the vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] is [tex]\(45^\circ\)[/tex].
So, the correct answer is:
2) [tex]\(45^\circ\)[/tex]
1. Dot Product:
The dot product of two vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] is given by:
[tex]\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \][/tex]
where [tex]\(\theta\)[/tex] is the angle between the vectors, and [tex]\(|\vec{A}|\)[/tex] and [tex]\(|\vec{B}|\)[/tex] are the magnitudes of [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] respectively.
2. Cross Product:
The magnitude of the cross product of [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] is given by:
[tex]\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \][/tex]
3. Given Condition:
According to the problem, we have:
[tex]\[ \vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}| \][/tex]
4. Substitute the Dot Product and Cross Product Formulas:
Substitute the expressions for the dot product and magnitude of the cross product:
[tex]\[ |\vec{A}| |\vec{B}| \cos \theta = |\vec{A}| |\vec{B}| \sin \theta \][/tex]
5. Simplify the Equation:
If both [tex]\(|\vec{A}|\)[/tex] and [tex]\(|\vec{B}|\)[/tex] are non-zero, we can divide both sides of the equation by [tex]\(|\vec{A}| |\vec{B}|\)[/tex]:
[tex]\[ \cos \theta = \sin \theta \][/tex]
6. Solve for [tex]\(\theta\)[/tex]:
Recall the trigonometric identity that [tex]\(\cos \theta = \sin \theta\)[/tex] when [tex]\(\theta\)[/tex] is 45 degrees. Therefore:
[tex]\[ \tan \theta = 1 \][/tex]
The angle whose tangent is 1 is:
[tex]\[ \theta = 45^\circ \][/tex]
Therefore, the angle between the vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] is [tex]\(45^\circ\)[/tex].
So, the correct answer is:
2) [tex]\(45^\circ\)[/tex]