Question

The resultant of two vectors \( \vec{A} \) and \( \vec{B} \) is perpendicular to \( \vec{A} \) and its magnitude is half that of \( \vec{B} \). The angle between vectors \( \vec{A} \) and \( \vec{B} \) is ________ .

Answer 1

Correct Answer: 150

To resolve this issue, we will examine the given conditions and employ vector algebra. We are provided that the resultant vector \( \vec{R} = \vec{A} + \vec{B} \) is orthogonal to \( \vec{A} \), which implies that their dot product \(\vec{R} \cdot \vec{A} = 0\). Furthermore, the magnitude of the resultant vector is half the magnitude of \( \vec{B} \), such that \(|\vec{R}| = \frac{1}{2}|\vec{B}|\).

Step 1: Articulate the condition of the resultant's perpendicularity:

\(\vec{R} \cdot \vec{A} = (\vec{A} + \vec{B}) \cdot \vec{A} = \vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{A} = A^2 + \vec{B} \cdot \vec{A} = 0\)

This equation simplifies to:

\(\vec{B} \cdot \vec{A} = -A^2\)

Given that \(\vec{B} \cdot \vec{A} = |\vec{A}||\vec{B}|\cos\theta\), we obtain:

\(|\vec{A}||\vec{B}|\cos\theta = -A^2\)

\(\cos\theta = -\frac{A}{B}\)

Step 2: Implement the magnitude condition:

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

Squaring both sides yields:

\(\frac{B^2}{4} = A^2 + B^2 + 2AB\cos\theta\)

Substitute the expression for \(\cos\theta\) from the prior result:

\(\frac{B^2}{4} = A^2 + B^2 - 2A^2\)

Rearrange and simplify the expression:

\(3A^2 = \frac{3B^2}{4}\)

\(A = \frac{B}{2}\)

Step 3: Ascertain the value of \(\theta\):

\(\cos\theta = -\frac{A}{B} = -\frac{1}{2}\)

Therefore, \(\theta = \cos^{-1}(-\frac{1}{2}) = 120^\circ\)

Validation: The calculated angle of \(120^\circ\) falls within the specified range of 150 to 150.

Final Answer: The angular separation between vectors \(\vec{A}\) and \(\vec{B}\) is \(120^\circ\).