If a+b+c=0 and |a|=5, |b|=3 and |c|=7,then th | vedclass

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(B) Given,$a+b+c=0$ and $|a|=5, |b|=3, |c|=7$.
Since $a+b+c=0$,we have $a+b=-c$.
Squaring both sides,we get $|a+b|^2 = |-c|^2$.
This implies $|a|^2

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$\frac{\pi}{3}$

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(B) Given,$a+b+c=0$ and $|a|=5, |b|=3, |c|=7$.
Since $a+b+c=0$,we have $a+b=-c$.
Squaring both sides,we get $|a+b|^2 = |-c|^2$.
This implies $|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$.
Using the definition of the dot product,$a \cdot b = |a||b| \cos 	heta$,where $	heta$ is the angle between $a$ and $b$.
Substituting the given values: $(5)^2 + (3)^2 + 2(5)(3) \cos 	heta = (7)^2$.
$25 + 9 + 30 \cos 	heta = 49$.
$34 + 30 \cos 	heta = 49$.
$30 \cos 	heta = 49 - 34 = 15$.
$\cos 	heta = \frac{15}{30} = \frac{1}{2}$.
Since $\cos 	heta = \frac{1}{2}$,we have $	heta = 60^{\circ}$ or $	heta = \frac{\pi}{3}$ radians.

If $a+b+c=0$ and $|a|=5, |b|=3$ and $|c|=7$,then the angle between $a$ and $b$ is

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