If the sum of two unit vectors is a unit vect | vedclass

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(B) Let $\hat{n}_1$ and $\hat{n}_2$ be two unit vectors. The magnitude of their sum is given by $|\hat{n}_1 + \hat{n}_2|^2 = |\hat{n}_1|^2 + |\hat{n}_

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$\sqrt{3}$

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(B) Let $\hat{n}_1$ and $\hat{n}_2$ be two unit vectors. The magnitude of their sum is given by $|\hat{n}_1 + \hat{n}_2|^2 = |\hat{n}_1|^2 + |\hat{n}_2|^2 + 2|\hat{n}_1||\hat{n}_2|\cos	heta$.Since the sum is a unit vector,$|\hat{n}_1 + \hat{n}_2| = 1$. Substituting the magnitudes $|\hat{n}_1| = 1$ and $|\hat{n}_2| = 1$,we get $1^2 = 1^2 + 1^2 + 2(1)(1)\cos	heta$.$1 = 2 + 2\cos	heta \implies 2\cos	heta = -1 \implies \cos	heta = -1/2$.This implies $	heta = 120^\circ$.The magnitude of the difference vector is given by $|\hat{n}_1 - \hat{n}_2| = \sqrt{|\hat{n}_1|^2 + |\hat{n}_2|^2 - 2|\hat{n}_1||\hat{n}_2|\cos	heta}$.Substituting the values,$|\hat{n}_1 - \hat{n}_2| = \sqrt{1^2 + 1^2 - 2(1)(1)(-1/2)} = \sqrt{1 + 1 + 1} = \sqrt{3}$.

If the sum of two unit vectors is a unit vector,then the magnitude of their difference is

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