Question

If \(\vec{a},\vec{b},\vec{c}\) are unit vectors satisfying \(\vec{a}-(\sqrt{3})\vec{b}+\vec{c}=\vec{0}\), then the angle between \(\vec{a}\) and \(\vec{c}\) is...

• A. \(\frac{\pi }{6}\)
• B. \(\frac{\pi }{3}\)
• C. \(\frac{\pi }{4}\)
• D. \(\frac{\pi }{2}\)

Hint:

Rearrange to a + c = root 3 times b and take the square of the magnitude.

Answer 1

The Correct Option is B

Step 1: Geometric View
$\vec{a}+\vec{c}$ is the diagonal of a rhombus with unit sides, and its length is $\sqrt{3}$.

Step 2: Use the Diagonal Formula
For a rhombus of side $1$ with angle $\theta$ between sides, the diagonal is $2\cos\dfrac{\theta}{2}$. So $2\cos\dfrac{\theta}{2}=\sqrt{3}$.

Step 3: Solve
$\cos\dfrac{\theta}{2}=\dfrac{\sqrt{3}}{2}\Rightarrow\dfrac{\theta}{2}=\dfrac{\pi}{6}\Rightarrow\theta=\dfrac{\pi}{3}$.

Final Answer:
The angle works out to pi/3. \[ \boxed{\text{(B) }\dfrac{\pi}{3}} \]