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(D) Given that $a, b,$ and $c$ are unit vectors,so $|a| = |b| = |c| = 1.$The given condition is $a + b - c = 0,$ which implies $a + b = c.$Taking the

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$2\pi /3$

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(D) Given that $a, b,$ and $c$ are unit vectors,so $|a| = |b| = |c| = 1.$The given condition is $a + b - c = 0,$ which implies $a + b = c.$Taking the dot product of both sides with themselves:$(a + b) \cdot (a + b) = c \cdot c$Expanding the dot product:$a \cdot a + b \cdot b + 2(a \cdot b) = |c|^2$Since $a \cdot a = |a|^2 = 1,$ $b \cdot b = |b|^2 = 1,$ and $c \cdot c = |c|^2 = 1,$$1 + 1 + 2|a||b|\cos 	heta = 1,$where $	heta$ is the angle between $a$ and $b.$$2 + 2(1)(1)\cos 	heta = 1$$2\cos 	heta = 1 - 2$$2\cos 	heta = -1$$\cos 	heta = -\frac{1}{2}$Since $\cos 	heta = -\frac{1}{2},$ the angle $	heta = \frac{2\pi}{3}.$

If $a, b$ and $c$ are unit vectors such that $a + b - c = 0,$ then the angle between $a$ and $b$ is

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