Question

A position vector ^ of magnitude 10 units makes an angle 53° with a
position vector B of magnitude 6 units. Find the magnitude of position
vector AB.​

Answer 1

Question:

A position vector A of magnitude 10 units makes an angle 53° with a

position vector B of magnitude 6 units. Find the magnitude of position

vector AB.​

Solution:

Magnitude of position vector AB(resultant vector),

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

$R = \sqrt{10^{2} + 6^{2} - 2(10)(6) cos 53^\circ }$

$R = \sqrt{100 + 36 - 120 * \frac{3}{5} }$

$R = \sqrt{100 + 36 - 24 * 3 }$

$R = \sqrt{100 + 36 - 72 }$

$R = \sqrt{64 }$

$R = 8$

The magnitude of position vector AB is 8 units.