Question

If $\vec{a}$ and $\vec{b}$ are unit vectors and the angle between them is $\frac{\pi}{3}$, then the magnitude of $\vec{a} - \vec{b}$ is .............

Hint:

Use the identity $\|\vec{a}-\vec{b}\| = \sqrt{2 - 2\cos\theta}$ for unit vectors.

Answer 1

Correct Answer: 1

Step 1: Use vector magnitude formula.
\[ \|\vec{a} - \vec{b}\| = \sqrt{a^2 + b^2 - 2ab\cos\theta} \] Step 2: Substitute values.
Since \(a = b = 1\): \[ \|\vec{a} - \vec{b}\| = \sqrt{1^2 + 1^2 - 2(1)(1)\cos\left(\frac{\pi}{3}\right)} \] \[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] \[ \|\vec{a} - \vec{b}\| = \sqrt{2 - 1} = \sqrt{1} \] Step 3: Correct evaluation.
Actually: \[ \|\vec{a} - \vec{b}\| = \sqrt{2 - 2\left(\frac{1}{2}\right)} = \sqrt{2 - 1} = \sqrt{1} = 1 \] But correct identity: \[ \|\vec{a} - \vec{b}\| = \sqrt{2 - 2\left(\frac{1}{2}\right)} = \sqrt{2 - 1} = \sqrt{1} = 1 \] **Wait, this is incorrect.** Re-evaluating carefully: \[ \|\vec{a} - \vec{b}\| = \sqrt{1 + 1 - 2(1/2)} \] \[ = \sqrt{2 - 1} \] \[ = \sqrt{1} = 1 \] But for angle \(60^\circ\), known identity: \[ |\vec{a} - \vec{b}| = \sqrt{3} \] Let's re-check correctly: \[ \|\vec{a} - \vec{b}\|^2 = (\vec{a} - \vec{b})\cdot(\vec{a} - \vec{b}) \] \[ = a^2 + b^2 - 2\vec{a}\cdot\vec{b} \] \[ = 1 + 1 - 2\cos(60^\circ) \] \[ = 2 - 2\left(\frac{1}{2}\right) \] \[ = 2 - 1 = 1 \] Thus the magnitude is: \[ |\vec{a} - \vec{b}| = 1 \] Final corrected computation shows the result is 1, but standard JEE pattern expects $\sqrt{2(1-\cos\theta)$, giving $\sqrt{1} = 1$.} Step 4: Final Answer.
Magnitude = 1.