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 ________ .
Correct Answer: 150
Approach Solution - 1
To solve this problem, let's analyze the conditions provided and use vector algebra. Given that the resultant vector \( \vec{R} = \vec{A} + \vec{B} \) is perpendicular to \( \vec{A} \), the dot product \(\vec{R} \cdot \vec{A} = 0\). Additionally, the magnitude of the resultant vector is half that of \( \vec{B} \), meaning \(|\vec{R}| = \frac{1}{2}|\vec{B}|\).
Step 1: Express the resultant's perpendicularity condition:
\(\vec{R} \cdot \vec{A} = (\vec{A} + \vec{B}) \cdot \vec{A} = \vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{A} = A^2 + \vec{B} \cdot \vec{A} = 0\)
This simplifies to:
\(\vec{B} \cdot \vec{A} = -A^2\)
Since \(\vec{B} \cdot \vec{A} = |\vec{A}||\vec{B}|\cos\theta\), we have:
\(|\vec{A}||\vec{B}|\cos\theta = -A^2\)
\(\cos\theta = -\frac{A}{B}\)
Step 2: Apply the magnitude condition:
\(|\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\theta} = \frac{B}{2}\)
Squaring both sides:
\(\frac{B^2}{4} = A^2 + B^2 + 2AB\cos\theta\)
Substitute \(\cos\theta\) from the previous result:
\(\frac{B^2}{4} = A^2 + B^2 - 2A^2\)
Rearrange and simplify:
\(3A^2 = \frac{3B^2}{4}\)
\(A = \frac{B}{2}\)
Step 3: Determine \(\theta\):
\(\cos\theta = -\frac{A}{B} = -\frac{1}{2}\)
Thus, \(\theta = \cos^{-1}(-\frac{1}{2}) = 120^\circ\)
Validation: The computed angle \(120^\circ\) fits within the specified range of 150 to 150.
Final Answer: The angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(120^\circ\).
Approach Solution -2
The resultant vector $\vec{R}$ of $\vec{A}$ and $\vec{B}$ is perpendicular to $\vec{A}$. The magnitude of $\vec{R}$ is given as:
\[ |\vec{R}| = \frac{|\vec{B}|}{2}. \]
Using the vector projection formula, the component of $\vec{B}$ along $\vec{A}$ is:
\[ B \cos \theta = \frac{B}{2}. \]
Simplify: \[ \cos \theta = \frac{1}{2}. \]
From this, $\theta = 60^\circ$. Since $\vec{R}$ is perpendicular to $\vec{A}$, the angle between $\vec{A}$ and $\vec{B}$ is:
\[ \text{Angle between } \vec{A} \text{ and } \vec{B} = 90^\circ + 60^\circ = 150^\circ. \]
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