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3-1.Vectors
easy
The modulus of the vector product of two vectors is $\frac{1}{\sqrt{3}}$ times their scalar product. The angle between vectors is
A$\frac{\pi}{6}$
B$\frac{\pi}{2}$
C$\frac{\pi}{4}$
D$\frac{\pi}{3}$
Solution
(a)
$A B \sin \theta=\frac{1}{\sqrt{3}} A B \cos \theta$
$\therefore \tan \theta=\frac{1}{\sqrt{3}} \text { or } \theta=30^{\circ}$
$A B \sin \theta=\frac{1}{\sqrt{3}} A B \cos \theta$
$\therefore \tan \theta=\frac{1}{\sqrt{3}} \text { or } \theta=30^{\circ}$
Standard 11
Physics
