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

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(D) Given,$a+b+c=0$.This implies $a+b=-c$.Squaring both sides,we get $(a+b)^2 = (-c)^2$.$|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$.Substituting the given

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$60$

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(D) Given,$a+b+c=0$.This implies $a+b=-c$.Squaring both sides,we get $(a+b)^2 = (-c)^2$.$|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$.Substituting the given magnitudes $|a|=3, |b|=5, |c|=7$:$3^2 + 5^2 + 2|a||b| \cos 	heta = 7^2$.$9 + 25 + 2(3)(5) \cos 	heta = 49$.$34 + 30 \cos 	heta = 49$.$30 \cos 	heta = 49 - 34$.$30 \cos 	heta = 15$.$\cos 	heta = \frac{15}{30} = \frac{1}{2}$.Therefore,$	heta = \cos^{-1}(\frac{1}{2}) = 60^{\circ}$.

If $a+b+c=0$ and $|a|=3, |b|=5, |c|=7$,then the angle between $a$ and $b$ is ........ (in $^{\circ}$)

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