If overrightarrow{a} cdot overrightarrow{b} = | vedclass

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Question

(B) We know that the dot product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\overrightarrow{a} \cdot \overrightarrow{b}

Vedclass · Accepted Answer

$180$

Vedclass · Answer

(B) We know that the dot product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}| \cos 	heta$,where $	heta$ is the angle between them.Given that $\overrightarrow{a} \cdot \overrightarrow{b} = -|\overrightarrow{a}||\overrightarrow{b}|$.Substituting the formula,we get $|\overrightarrow{a}||\overrightarrow{b}| \cos 	heta = -|\overrightarrow{a}||\overrightarrow{b}|$.Assuming $\overrightarrow{a}$ and $\overrightarrow{b}$ are non-zero vectors,we can divide both sides by $|\overrightarrow{a}||\overrightarrow{b}|$.This gives $\cos 	heta = -1$.The value of $	heta$ for which $\cos 	heta = -1$ is $	heta = 180^{\circ}$.

If $\overrightarrow{a} \cdot \overrightarrow{b} = -|\overrightarrow{a}||\overrightarrow{b}|$,then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is (in $^{\circ}$)

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