Question

Vector A has a magnitude of 2 units and makes an angle of ${30}^{\circ }$ with the positive x- axis, Vector B has a magnitude of 6 units and makes an angle of ${30}^{\circ }$ with the y- axis(clockwise). Find the magnitude of the resultant of Vector A & -B and also the angle that it makes with Vector A.

• A. Magnitude =
• B. Magnitude =
• C. Magnitude =
• D. Magnitude =

Answer 1

The correct option is C

Magnitude =

Given:

Using parallelogram law

Now we know magnitude of resultant is give by the formula

$\sqrt{a^2+b^2+2abcos\theta }$

So we have

$\sqrt{2^2+6^2+24cos{150}^{\circ }}=\sqrt{28}$

Now we know to find the angle that it makes with the A vector we have the formula

$tan\theta =\frac{bsin\theta }{(a+bcos\theta )}$

$tan\theta =\frac{6sin{150}^{\circ }}{(2+6cos{150}^{\circ })}=\frac{-3\sqrt{3}}{(2-3)}=3\sqrt{3}$