Question

The resultant of three vectors 1, 2 and 3 units whose directions are those of the sides of an eauilateral triangle taken in an order is at an angle of

• A. ${30}^{\circ }$ with the first vector
• B. ${15}^{\circ }$ with the first vector
• C. ${100}^{\circ }$ with the first vector
• D. ${150}^{\circ }$ with the first vector

Answer 1

Answer: (D)

${150}^{\circ }$ with the first vector

Analytically :

$R_x=1+2cos{120}^0+3cos{240}^0=-3/2$

$R_y=2sin{120}^0+3sin{240}^0$

$=2\times \frac{\sqrt{3}}{2}+3\times (-\frac{\sqrt{3}}{2})=-\frac{\sqrt{3}}{2}$

$tan\theta =\frac{R_y}{R_x}=\frac{-\frac{\sqrt{3}}{2}}{-\frac{3}{2}}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\Rightarrow \theta ={30}^0$

Hence, angle with first vector is $180-\theta ={150}^0$

$AB=1,BC=2,CD=3$

$\overrightarrow{R}=\overrightarrow{AD}$ is resultant, clearly angle between $\overrightarrow{R}$ and $\overrightarrow{AB}$ is ${150}^0$