Question

Vectors $\overrightarrow{C}$ and $\overrightarrow{D}$ have magnitudes of 3 and 4 units respectively. What is the angle between the directions of $\overrightarrow{C}$ and $\overrightarrow{D}$ if the magnitude of vector product $\overrightarrow{C}\times \overrightarrow{D}$ is x and y respectively?

(x) 0 (y) 12 units

(i) 0 (ii)$\frac{\pi }{2}$ (iii)$\frac{\pi }{3}$ (iv) $\frac{\pi }{4}$

• A. (x) - (iv); (y) - (iii)
• B. (x) - (i); (y) - (ii)
• C. (x) - (ii); (y) - (i)
• D. (x) - (i); (y) - (iv)

Answer 1

The correct option is B

(x) - (i); (y) - (ii)

Cross product magnitude is given by
$|\overrightarrow{R}|=|\overrightarrow{C}\times \overrightarrow{D}|=|C||D|sin\theta$
where $\theta$ is the angle between $\overrightarrow{C}$ and $\overrightarrow{D}$
$\therefore$ R=12 sin $\theta$ (as |C|=3 and |D|=4)
$\Rightarrow ifR=0,\theta =0andifR=12,sin\theta =1\Rightarrow \theta ={90}^{\circ }or\frac{\pi }{2}$