Question

Vector A has a magnitude of 2 units and makes an angle of $0^{\circ }$ with the positive x- axis, Vector B has a magnitude of 6 units and makes an angle of ${30}^{\circ }$
(clock-wise) with the positive y- axis. Find the magnitude of the resultant of Vector A& B and also the angle that it makes with Vector A.

• A. Magnitude = $\sqrt{52}$ $Tan\theta =3\sqrt{3}$
• B. Magnitude = $\sqrt{40}$ $Tan\theta =\frac{3}{2+3\sqrt{3}}$
• C. Magnitude = $\sqrt{40}$ $Tan\theta =3\sqrt{3}$
• D. Magnitude = $\sqrt{52}$ $Tan\theta =\frac{3\sqrt{3}}{5}$

Answer 1

The correct option is D

Magnitude = $\sqrt{52}$ $Tan\theta =\frac{3\sqrt{3}}{5}$

By applying parallelogram law this can be made to look like

Now we know magnitude of resultant is given by the formula

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

So we hace

$\sqrt{2^2+6^2+24cos{60}^{\circ }}=\sqrt{52}$

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

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

$tan\theta =\frac{6sin{60}^{\circ }}{(2+6cos{60}^{\circ })}=\frac{3\sqrt{3}}{(2+3)}$