© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: State the Given Condition
We are given that the magnitude of the sum of the vectors is equal to the magnitude of their difference:
$|\overrightarrow{A} + \overrightarrow{B}| = |\overrightarrow{A} - \overrightarrow{B}|.$
Step 2: Square Both Sides of the Equality
When two magnitudes are equal, the squares of these magnitudes are also equal. Thus:
$|\overrightarrow{A} + \overrightarrow{B}|^2 = |\overrightarrow{A} - \overrightarrow{B}|^2.$
Step 3: Expand the Magnitudes in Terms of Dot Products
Recall that for any vectors $\overrightarrow{X}$ and $\overrightarrow{Y}$,
$|\overrightarrow{X} + \overrightarrow{Y}|^2 = |\overrightarrow{X}|^2 + |\overrightarrow{Y}|^2 + 2\,\overrightarrow{X} \cdot \overrightarrow{Y}.$
Similarly,
$|\overrightarrow{X} - \overrightarrow{Y}|^2 = |\overrightarrow{X}|^2 + |\overrightarrow{Y}|^2 - 2\,\overrightarrow{X} \cdot \overrightarrow{Y}.$
Thus, we can write:
$
|\overrightarrow{A} + \overrightarrow{B}|^2
= A^2 + B^2 + 2\,\overrightarrow{A} \cdot \overrightarrow{B}
= A^2 + B^2 + 2AB\,\cos \theta,
$
and
$
|\overrightarrow{A} - \overrightarrow{B}|^2
= A^2 + B^2 - 2\,\overrightarrow{A} \cdot \overrightarrow{B}
= A^2 + B^2 - 2AB\,\cos \theta.
$
Step 4: Equate the Two Expressions and Simplify
Since both squared magnitudes are equal:
$A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 - 2AB \cos \theta.$
Subtracting $A^2 + B^2$ from both sides gives:
$2AB \cos \theta = -2AB \cos \theta.$
Therefore,
$4AB \cos \theta = 0 \quad \Longrightarrow \quad \cos \theta = 0.$
Step 5: Evaluate the Angle
From $\cos \theta = 0,$ we find that $\theta = 90^\circ.$ Hence, the vectors are perpendicular to each other.
Answer
The angle between the two vectors is $90^\circ$.
