Question

Find the magnitude of the vectors \(\vec a\) and \(\vec b\), if having equal magnitudes and angle \(60^\circ\) between them and their scalar product is \(\dfrac12\).

Hint:

Use \(\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta\) with \(|\vec a|=|\vec b|\).

Answer 1

Step 1: Using |a+b|^2 as a cross-check:
\(|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\,\vec a\cdot\vec b=k^2+k^2+2\left(\dfrac12\right)=2k^2+1\).

Step 2: Independently solving from the dot product directly:
Since \(\vec a\cdot\vec b=k^2\cos60^\circ=\dfrac{k^2}{2}=\dfrac12\), the same equation \(k^2=1\) results regardless of which identity is used, giving \(k=1\).

Final Answer:
So \(|\vec a|=|\vec b|=\boxed{1}\), consistent with the direct method.