If the magnitudes of vectors A→,B→ and C→ and 12,5 and 13 units respectively and A→+B→=C→, the angle between vectors  A→ and B→ is:

# If the magnitudes of vectors  and 12,5 and 13 units respectively and $\stackrel{\to }{\mathrm{A}}+\stackrel{\to }{\mathrm{B}}=\stackrel{\to }{\mathrm{C}}$, the angle between vectors   is:

1. A

0

2. B

$\mathrm{\pi }$

3. C

$\mathrm{\pi }/2$

4. D

$\mathrm{\pi }/4$

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### Solution:

$\stackrel{\to }{\mathrm{A}}+\stackrel{\to }{\mathrm{B}}=\stackrel{\to }{\mathrm{C}}⇒\left|\stackrel{\to }{\mathrm{A}}+\stackrel{\to }{\mathrm{B}}\right|=\left|\stackrel{\to }{\mathrm{C}}\right|$
Let angle between $\stackrel{\to }{\mathrm{A}}$and $\stackrel{\to }{\mathrm{B}}$ be $\mathrm{\theta }$ then A2 + B2 + 2ABcos$\mathrm{\theta }$ = $\mathrm{\pi }$/2$⇒$
$⇒$ 122 + 52 + 2(12)(5)cos$\mathrm{\theta }$ = 132
$⇒$cos$\mathrm{\theta }$ = 0$⇒$$\mathrm{\theta }$=$\frac{\mathrm{\pi }}{2}\mathrm{rad}$
OR
$\because$132 =122 + 52 i.e.,C2 = A2 + B2
So,$\stackrel{\to }{\mathrm{A}}\perp \stackrel{\to }{\mathrm{B}}$

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