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3-1.Vectors
medium
If $| A |=2$ and $| B |=4$ and angle between them is $60^{\circ}$, then $| A - B |$ is
A$\sqrt{13}$
B$3 \sqrt{3}$
C$\sqrt{3}$
D$2 \sqrt{3}$
Solution
(d)
$| A – B | =\sqrt{A^2+B^2-2 A B \cos \theta}$
$=\sqrt{4+16-2 \times 2 \times 4 \times \frac{1}{2}}$
$=\sqrt{12}=2 \sqrt{3}$
$| A – B | =\sqrt{A^2+B^2-2 A B \cos \theta}$
$=\sqrt{4+16-2 \times 2 \times 4 \times \frac{1}{2}}$
$=\sqrt{12}=2 \sqrt{3}$
Standard 11
Physics