Question

The magnitude of the sum of the two vectors is equal to the difference of their magnitudes, the angle between the vectors is,

• A. 0$^\circ$
• B. 45$^\circ$
• C. 90$^\circ$
• D. 180$^\circ$

Answer 1

The Correct Option is C

Magnitude of Vector Sum and Difference 

Let the two vectors be \( \vec{A} \) and \( \vec{B} \) with magnitudes \( A \) and \( B \), respectively. Then, the magnitude of their sum is given by:

\(|\overrightarrow{ A } + \overrightarrow{ B }| = \sqrt{A^2 + B^2 + 2AB \cos \theta} \) (\)

Where \( \theta \) is the angle between the vectors.

The magnitude of their difference is given by:

\(|\overrightarrow{ A } - \overrightarrow{ B }| = \sqrt{A^2 + B^2 - 2AB \cos \theta} \) (\)

Now, according to the question, we are given that:

\(|\overrightarrow{ A } + \overrightarrow{ B }| = |\overrightarrow{ A } - \overrightarrow{ B }|\)

From this, we have:

\(A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 - 2AB \cos \theta\)

Simplifying the above equation:

\(4AB \cos \theta = 0\)

This gives us:

\(\cos \theta = 0\)

Thus, the angle between the two vectors is:

\(\theta = 90^\circ\)

Conclusion:

The angle between the vectors \( \vec{A} \) and \( \vec{B} \) is \( 90^\circ \), meaning the vectors are perpendicular to each other.