Question

Find the magnitude of two vectors `veca and vecb`, having the same magnitude and such that the angle between them is 60° and their scalar product is `1/2`.

Answer 1

We have, `theta = 60^circ, veca xx vecb = 1/2, |veca| = |vecb|`

If the angle between vectors `veca, vecb` is θ, then

Now, `costheta = (veca xx vecb)/(|veca||vecb|)`

`cos 60^circ = (1/2)/|veca|^2; 1/2 = 1/(2|veca|^2)`

`|veca|^2 = 1; |veca| = 1`

`|veca| = 1, |vecb| = 1`

Answer 2

Magnitude of two vectors `veca and vecb` is same

`|veca| = |vecb|`

`veca.vecb = |veca| |vecb| costheta`, θ is the angle between `veca and vecb`

Given `theta = 60^\circ and veca.vecb = 1/2`

`veca.vecb = |veca| |vecb| costheta`

`veca.vecb = |veca| |veca| cos60^\circ`

`1/2 = |veca|^2 xx 1/2`

`|veca|^2 = 1`

`|veca| = +- 1`

Since magnitude of a vector is not negative

so `|veca| = 1`

`|veca| = |vecb| = 1`