If a and b are unit vectors such that a+b is | vedclass

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(D) Given that $a$ and $b$ are unit vectors,we have $|a| = 1$ and $|b| = 1$. Also,$a+b$ is a unit vector,so $|a+b| = 1$. Squaring both sides,we get $|

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) Given that $a$ and $b$ are unit vectors,we have $|a| = 1$ and $|b| = 1$. Also,$a+b$ is a unit vector,so $|a+b| = 1$. Squaring both sides,we get $|a+b|^2 = 1^2 = 1$. Using the property $|a+b|^2 = |a|^2 + |b|^2 + 2(a \cdot b)$,we have: $1^2 + 1^2 + 2(a \cdot b) = 1$ $1 + 1 + 2(a \cdot b) = 1$ $2 + 2(a \cdot b) = 1$ $2(a \cdot b) = -1$ $a \cdot b = -\frac{1}{2}$. Since $a \cdot b = |a||b| \cos 	heta$,where $	heta$ is the angle between $a$ and $b$: $(1)(1) \cos 	heta = -\frac{1}{2}$ $\cos 	heta = -\frac{1}{2}$. Therefore,$	heta = 120^{\circ}$. Hence,option $D$ is correct.

If $a$ and $b$ are unit vectors such that $a+b$ is also a unit vector,then the angle between $a$ and $b$ is . . . . . . (in $^{\circ}$)

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