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Question: If two equal vectors $\overline{A}$ and $\overline{B}$ are inclined at an angle $\theta$ and each of...

If two equal vectors A‾\overline{A} and B‾\overline{B} are inclined at an angle θ\theta and each of magnitude is f, then ∣A‾−B‾∣|\overline{A}-\overline{B}|

A

2fcos⁡(θ2)2f \cos (\frac{\theta}{2})

B

2fsin⁡(θ2)2f \sin (\frac{\theta}{2})

C

f2(1−sin⁡ϕ)f\sqrt{2(1-\sin\phi)}

D

f2(1+sin⁡ϕ)f\sqrt{2(1+\sin\phi)}

Answer

2fsin⁡(θ2)2f \sin (\frac{\theta}{2})

Explanation

Solution

Let the two equal vectors be A‾\overline{A} and B‾\overline{B}. The magnitude of each vector is given as ff, so ∣A‾∣=f|\overline{A}| = f and ∣B‾∣=f|\overline{B}| = f. The angle between the two vectors is θ\theta. We need to find the magnitude of the difference between the two vectors, ∣A‾−B‾∣|\overline{A} - \overline{B}|.

The magnitude of the difference of two vectors A‾\overline{A} and B‾\overline{B} is given by the formula: ∣A‾−B‾∣=∣A‾∣2+∣B‾∣2−2∣A‾∣∣B‾∣cos⁡θ|\overline{A} - \overline{B}| = \sqrt{|\overline{A}|^2 + |\overline{B}|^2 - 2|\overline{A}||\overline{B}|\cos\theta}

Substitute the given magnitudes ∣A‾∣=f|\overline{A}| = f and ∣B‾∣=f|\overline{B}| = f into the formula: ∣A‾−B‾∣=f2+f2−2(f)(f)cos⁡θ|\overline{A} - \overline{B}| = \sqrt{f^2 + f^2 - 2(f)(f)\cos\theta} ∣A‾−B‾∣=2f2−2f2cos⁡θ|\overline{A} - \overline{B}| = \sqrt{2f^2 - 2f^2\cos\theta}

Factor out 2f22f^2 from the terms under the square root: ∣A‾−B‾∣=2f2(1−cos⁡θ)|\overline{A} - \overline{B}| = \sqrt{2f^2(1 - \cos\theta)}

Now, use the trigonometric identity 1−cos⁡θ=2sin⁡2(θ2)1 - \cos\theta = 2\sin^2(\frac{\theta}{2}). Substitute this identity into the expression: ∣A‾−B‾∣=2f2(2sin⁡2(θ2))|\overline{A} - \overline{B}| = \sqrt{2f^2(2\sin^2(\frac{\theta}{2}))} ∣A‾−B‾∣=4f2sin⁡2(θ2)|\overline{A} - \overline{B}| = \sqrt{4f^2\sin^2(\frac{\theta}{2})}

Taking the square root, we get: ∣A‾−B‾∣=(2fsin⁡(θ2))2|\overline{A} - \overline{B}| = \sqrt{(2f\sin(\frac{\theta}{2}))^2} ∣A‾−B‾∣=∣2fsin⁡(θ2)∣|\overline{A} - \overline{B}| = |2f\sin(\frac{\theta}{2})|

Since ff is a magnitude, f≥0f \ge 0. The angle θ\theta between two vectors is typically considered in the range 0≤θ≤π0 \le \theta \le \pi. For this range, 0≤θ2≤π20 \le \frac{\theta}{2} \le \frac{\pi}{2}, and sin⁡(θ2)≥0\sin(\frac{\theta}{2}) \ge 0. Therefore, ∣sin⁡(θ2)∣=sin⁡(θ2)|\sin(\frac{\theta}{2})| = \sin(\frac{\theta}{2}).

So, the magnitude of the difference is: ∣A‾−B‾∣=2fsin⁡(θ2)|\overline{A} - \overline{B}| = 2f\sin(\frac{\theta}{2})