Question

Angle between 2 vectors of magnitude 12 and 18 unit, when their resultant is 24 units

Answer 1

This should help.......

Answer 2

The angle between the units of vectors is $\bpld{\theta=\cos ^{-1}(0.64)}$

SOLUTION:

Let one of the vector be A

Magnitude of A = 12 unit

Let another vector be B

Magnitude of B = 18 unit.

The resultant vector is given as

$=\sqrt{A^{2}}+\sqrt{B^{2}}+\sqrt{2 A B \cos \theta}$

The resultant vector of A and B is given as 24 units.

We need to find the angleθ.

Substituting the values,

$\sqrt{\left(12^{2}+18^{2}+2 \times 12 \times 18 \times \cos \theta\right)}=\sqrt{24}30 \cos \theta=2 \sqrt{6}$

$\theta=\cos ^{-1}(0.64)$

Thus the ‘angle between the vectors’ of 12 units and 18 units will be cosine inverse of 0.64.