If overrightarrow{a} and overrightarrow{b} ar | vedclass

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(D) Given that $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors,so $|\overrightarrow{a}| = 1$ and $|\overrightarrow{b}| = 1$. We are giv

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$\sqrt{3}$

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(D) Given that $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors,so $|\overrightarrow{a}| = 1$ and $|\overrightarrow{b}| = 1$. We are given $|\overrightarrow{a} + \overrightarrow{b}| = 1$. Squaring both sides,we get $|\overrightarrow{a} + \overrightarrow{b}|^2 = 1^2$. Using the property $|\overrightarrow{x} + \overrightarrow{y}|^2 = |\overrightarrow{x}|^2 + |\overrightarrow{y}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b})$,we have: $|\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b}) = 1$. Substituting the values,$1^2 + 1^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b}) = 1$. $2 + 2(\overrightarrow{a} \cdot \overrightarrow{b}) = 1 \Rightarrow 2(\overrightarrow{a} \cdot \overrightarrow{b}) = -1$. Now,we need to find $|\overrightarrow{a} - \overrightarrow{b}|$. $|\overrightarrow{a} - \overrightarrow{b}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 - 2(\overrightarrow{a} \cdot \overrightarrow{b})$. Substituting the values,$|\overrightarrow{a} - \overrightarrow{b}|^2 = 1^2 + 1^2 - (-1) = 1 + 1 + 1 = 3$. Therefore,$|\overrightarrow{a} - \overrightarrow{b}| = \sqrt{3}$.

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors and $|\overrightarrow{a}+\overrightarrow{b}|=1$,then $|\overrightarrow{a}-\overrightarrow{b}|$ is equal to

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