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If ∣a⃗−r⃗∣=∣a⃗∣=∣r⃗∣=1|\vec{a} - \vec{r}| = |\vec{a}| = |\vec{r}| = 1, then angle between a⃗\vec{a} and r⃗\vec{r} is

Solution

✅ Correct Option: 1

Given ∣a⃗−r⃗∣=1|\vec{a} - \vec{r}| = 1, ∣a⃗∣=1|\vec{a}| = 1, and ∣r⃗∣=1|\vec{r}| = 1.

The magnitude of the difference between two vectors can be expressed as:

∣a⃗−r⃗∣2=∣a⃗∣2+∣r⃗∣2−2∣a⃗∣∣r⃗∣cos⁡θ|\vec{a} - \vec{r}|^2 = |\vec{a}|^2 + |\vec{r}|^2 - 2|\vec{a}||\vec{r}|\cos\theta

where θ\theta is the angle between the vectors.


Substituting the given values:

12=12+12−2(1)(1)cos⁡θ1^2 = 1^2 + 1^2 - 2(1)(1)\cos\theta

1=1+1−2cos⁡θ1 = 1 + 1 - 2\cos\theta

1=2−2cos⁡θ1 = 2 - 2\cos\theta


2cos⁡θ=2−12\cos\theta = 2 - 1

2cos⁡θ=12\cos\theta = 1

cos⁡θ=12\cos\theta = \frac{1}{2}


The angle whose cosine equals 12\frac{1}{2} is:

θ=π3\theta = \frac{\pi}{3}

Therefore, the angle between a⃗\vec{a} and r⃗\vec{r} is π3\frac{\pi}{3}.

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