Answer :
Certainly! Let's work through the problem step-by-step:
We have two vectors [tex]\( \mathbf{A} \)[/tex] and [tex]\( \mathbf{B} \)[/tex], and we need to determine the angle [tex]\( \theta \)[/tex] between them given that [tex]\( |\mathbf{A} + \mathbf{B}| = |\mathbf{A} - \mathbf{B}| \)[/tex].
### Step-by-Step Solution:
1. Magnitude of Vectors:
We need the magnitudes of [tex]\( \mathbf{A} + \mathbf{B} \)[/tex] and [tex]\( \mathbf{A} - \mathbf{B} \)[/tex]. The magnitude of a vector [tex]\( \mathbf{V} \)[/tex] is given by [tex]\( |\mathbf{V}| = \sqrt{\mathbf{V} \cdot \mathbf{V}} \)[/tex], where [tex]\( \cdot \)[/tex] denotes the dot product.
2. Magnitude of [tex]\( \mathbf{A} + \mathbf{B} \)[/tex]:
[tex]\[ |\mathbf{A} + \mathbf{B}| = \sqrt{(\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} + \mathbf{B})} \][/tex]
Expanding the dot product:
[tex]\[ (\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} + \mathbf{B}) = \mathbf{A} \cdot \mathbf{A} + 2 \mathbf{A} \cdot \mathbf{B} + \mathbf{B} \cdot \mathbf{B} \][/tex]
Thus,
[tex]\[ |\mathbf{A} + \mathbf{B}| = \sqrt{A^2 + 2 \mathbf{A} \cdot \mathbf{B} + B^2} \][/tex]
where [tex]\( A = |\mathbf{A}| \)[/tex] and [tex]\( B = |\mathbf{B}| \)[/tex].
3. Magnitude of [tex]\( \mathbf{A} - \mathbf{B} \)[/tex]:
[tex]\[ |\mathbf{A} - \mathbf{B}| = \sqrt{(\mathbf{A} - \mathbf{B}) \cdot (\mathbf{A} - \mathbf{B})} \][/tex]
Expanding the dot product:
[tex]\[ (\mathbf{A} - \mathbf{B}) \cdot (\mathbf{A} - \mathbf{B}) = \mathbf{A} \cdot \mathbf{A} - 2 \mathbf{A} \cdot \mathbf{B} + \mathbf{B} \cdot \mathbf{B} \][/tex]
Thus,
[tex]\[ |\mathbf{A} - \mathbf{B}| = \sqrt{A^2 - 2 \mathbf{A} \cdot \mathbf{B} + B^2} \][/tex]
4. Equating the Magnitudes:
Given [tex]\( |\mathbf{A} + \mathbf{B}| = |\mathbf{A} - \mathbf{B}| \)[/tex]:
[tex]\[ \sqrt{A^2 + 2 \mathbf{A} \cdot \mathbf{B} + B^2} = \sqrt{A^2 - 2 \mathbf{A} \cdot \mathbf{B} + B^2} \][/tex]
5. Removing the Square Roots:
Since both sides are non-negative, we can square both sides of the equation:
[tex]\[ A^2 + 2 \mathbf{A} \cdot \mathbf{B} + B^2 = A^2 - 2 \mathbf{A} \cdot \mathbf{B} + B^2 \][/tex]
6. Solving for the Dot Product:
Simplifying this, we get:
[tex]\[ A^2 + 2 \mathbf{A} \cdot \mathbf{B} + B^2 = A^2 - 2 \mathbf{A} \cdot \mathbf{B} + B^2 \][/tex]
[tex]\[ 2 \mathbf{A} \cdot \mathbf{B} = -2 \mathbf{A} \cdot \mathbf{B} \][/tex]
Adding [tex]\( 2 \mathbf{A} \cdot \mathbf{B} \)[/tex] to both sides:
[tex]\[ 4 \mathbf{A} \cdot \mathbf{B} = 0 \][/tex]
Thus,
[tex]\[ \mathbf{A} \cdot \mathbf{B} = 0 \][/tex]
7. Interpreting the Dot Product:
[tex]\[ \mathbf{A} \cdot \mathbf{B} = AB \cos \theta = 0 \][/tex]
Since [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are non-zero magnitudes, we have:
[tex]\[ \cos \theta = 0 \][/tex]
8. Finding the Angle [tex]\( \theta \)[/tex]:
The angle whose cosine is 0 is:
[tex]\[ \theta = 90^\circ \][/tex]
### Conclusion:
Therefore, given that [tex]\( |\mathbf{A} + \mathbf{B}| = |\mathbf{A} - \mathbf{B}| \)[/tex], the angle [tex]\( \theta \)[/tex] between vectors [tex]\( \mathbf{A} \)[/tex] and [tex]\( \mathbf{B} \)[/tex] is [tex]\( 90^\circ \)[/tex].
We have two vectors [tex]\( \mathbf{A} \)[/tex] and [tex]\( \mathbf{B} \)[/tex], and we need to determine the angle [tex]\( \theta \)[/tex] between them given that [tex]\( |\mathbf{A} + \mathbf{B}| = |\mathbf{A} - \mathbf{B}| \)[/tex].
### Step-by-Step Solution:
1. Magnitude of Vectors:
We need the magnitudes of [tex]\( \mathbf{A} + \mathbf{B} \)[/tex] and [tex]\( \mathbf{A} - \mathbf{B} \)[/tex]. The magnitude of a vector [tex]\( \mathbf{V} \)[/tex] is given by [tex]\( |\mathbf{V}| = \sqrt{\mathbf{V} \cdot \mathbf{V}} \)[/tex], where [tex]\( \cdot \)[/tex] denotes the dot product.
2. Magnitude of [tex]\( \mathbf{A} + \mathbf{B} \)[/tex]:
[tex]\[ |\mathbf{A} + \mathbf{B}| = \sqrt{(\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} + \mathbf{B})} \][/tex]
Expanding the dot product:
[tex]\[ (\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} + \mathbf{B}) = \mathbf{A} \cdot \mathbf{A} + 2 \mathbf{A} \cdot \mathbf{B} + \mathbf{B} \cdot \mathbf{B} \][/tex]
Thus,
[tex]\[ |\mathbf{A} + \mathbf{B}| = \sqrt{A^2 + 2 \mathbf{A} \cdot \mathbf{B} + B^2} \][/tex]
where [tex]\( A = |\mathbf{A}| \)[/tex] and [tex]\( B = |\mathbf{B}| \)[/tex].
3. Magnitude of [tex]\( \mathbf{A} - \mathbf{B} \)[/tex]:
[tex]\[ |\mathbf{A} - \mathbf{B}| = \sqrt{(\mathbf{A} - \mathbf{B}) \cdot (\mathbf{A} - \mathbf{B})} \][/tex]
Expanding the dot product:
[tex]\[ (\mathbf{A} - \mathbf{B}) \cdot (\mathbf{A} - \mathbf{B}) = \mathbf{A} \cdot \mathbf{A} - 2 \mathbf{A} \cdot \mathbf{B} + \mathbf{B} \cdot \mathbf{B} \][/tex]
Thus,
[tex]\[ |\mathbf{A} - \mathbf{B}| = \sqrt{A^2 - 2 \mathbf{A} \cdot \mathbf{B} + B^2} \][/tex]
4. Equating the Magnitudes:
Given [tex]\( |\mathbf{A} + \mathbf{B}| = |\mathbf{A} - \mathbf{B}| \)[/tex]:
[tex]\[ \sqrt{A^2 + 2 \mathbf{A} \cdot \mathbf{B} + B^2} = \sqrt{A^2 - 2 \mathbf{A} \cdot \mathbf{B} + B^2} \][/tex]
5. Removing the Square Roots:
Since both sides are non-negative, we can square both sides of the equation:
[tex]\[ A^2 + 2 \mathbf{A} \cdot \mathbf{B} + B^2 = A^2 - 2 \mathbf{A} \cdot \mathbf{B} + B^2 \][/tex]
6. Solving for the Dot Product:
Simplifying this, we get:
[tex]\[ A^2 + 2 \mathbf{A} \cdot \mathbf{B} + B^2 = A^2 - 2 \mathbf{A} \cdot \mathbf{B} + B^2 \][/tex]
[tex]\[ 2 \mathbf{A} \cdot \mathbf{B} = -2 \mathbf{A} \cdot \mathbf{B} \][/tex]
Adding [tex]\( 2 \mathbf{A} \cdot \mathbf{B} \)[/tex] to both sides:
[tex]\[ 4 \mathbf{A} \cdot \mathbf{B} = 0 \][/tex]
Thus,
[tex]\[ \mathbf{A} \cdot \mathbf{B} = 0 \][/tex]
7. Interpreting the Dot Product:
[tex]\[ \mathbf{A} \cdot \mathbf{B} = AB \cos \theta = 0 \][/tex]
Since [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are non-zero magnitudes, we have:
[tex]\[ \cos \theta = 0 \][/tex]
8. Finding the Angle [tex]\( \theta \)[/tex]:
The angle whose cosine is 0 is:
[tex]\[ \theta = 90^\circ \][/tex]
### Conclusion:
Therefore, given that [tex]\( |\mathbf{A} + \mathbf{B}| = |\mathbf{A} - \mathbf{B}| \)[/tex], the angle [tex]\( \theta \)[/tex] between vectors [tex]\( \mathbf{A} \)[/tex] and [tex]\( \mathbf{B} \)[/tex] is [tex]\( 90^\circ \)[/tex].
