MCQ
If ${(a \times b)^2} + {(a\,\,.\,\,b)^2} = 144$ and $|a|\, = 4,$ then $|b|\, = $
- A$16$
- B$8$
- ✓$3$
- D$12$
$\therefore \,\,\,144 = 16|b{|^2} \Rightarrow \,|b| = 3.$
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$x+y+\sqrt{3} z=0$
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$x+y+(\tan \theta) z=0$
has non-trivial solution. Then $\frac{120}{\pi} \sum_{\theta \in s} \theta$ is equal to