First slide

Vectors

Question

If P + Q = Rand IPI = IQI = $\sqrt{3}$ and IRI = 3, then the angle between P and Q is

Easy

A

$π / 4$
B

$π / 6$
C

$π / 3$
D

$π / 2$

Solution

Resultant, $R = \sqrt{P^{2} + Q^{2} + 2 P Q \cos θ}$
$3 = \sqrt{(\sqrt{3})^{2} + (\sqrt{3})^{2} + 2 (\sqrt{3}) (\sqrt{3}) \cos θ}$
Squaring both sides in the above equation, we get
9 = 3 + 3+ 6 cos $θ$
$\Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = 60 ° or \frac{π}{3}$

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