If |a| = 3, |b| = 4, |c| = 5 and a + b + c = | vedclass

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(D) Given that $a + b + c = 0$,we can write $a + b = -c$.Squaring both sides,we get $|a + b|^2 = |-c|^2$.Using the property $|x|^2 = x \cdot x$,we hav

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$\frac{\pi}{2}$

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(D) Given that $a + b + c = 0$,we can write $a + b = -c$.Squaring both sides,we get $|a + b|^2 = |-c|^2$.Using the property $|x|^2 = x \cdot x$,we have $(a + b) \cdot (a + b) = |c|^2$.Expanding the dot product,we get $|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$.Since $a \cdot b = |a||b| \cos 	heta$,where $	heta$ is the angle between $a$ and $b$,we have $|a|^2 + |b|^2 + 2|a||b| \cos 	heta = |c|^2$.Substituting the given values $|a| = 3, |b| = 4, |c| = 5$:$3^2 + 4^2 + 2(3)(4) \cos 	heta = 5^2$$9 + 16 + 24 \cos 	heta = 25$$25 + 24 \cos 	heta = 25$$24 \cos 	heta = 0$$\cos 	heta = 0$Therefore,$	heta = \frac{\pi}{2}$.

If $|a| = 3, |b| = 4, |c| = 5$ and $a + b + c = 0,$ then the angle between $a$ and $b$ is

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