If theta is the angle between the unit vector | vedclass

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(B) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.Since $a - \sqrt{2}b$ is a unit vector,its magnitude is $1$,so $|a - \sqrt{2}b|

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$\frac{\pi}{4}$

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(B) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.Since $a - \sqrt{2}b$ is a unit vector,its magnitude is $1$,so $|a - \sqrt{2}b| = 1$.Squaring both sides,we get $|a - \sqrt{2}b|^2 = 1^2$.Using the property $|u - v|^2 = |u|^2 + |v|^2 - 2(u \cdot v)$,we have:$|a|^2 + |\sqrt{2}b|^2 - 2(a \cdot \sqrt{2}b) = 1$.Substituting the values $|a| = 1$ and $|b| = 1$:$1^2 + 2|b|^2 - 2\sqrt{2}(a \cdot b) = 1$.$1 + 2(1) - 2\sqrt{2}(a \cdot b) = 1$.$3 - 2\sqrt{2}(a \cdot b) = 1$.$2\sqrt{2}(a \cdot b) = 2$.$a \cdot b = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.Since $a \cdot b = |a||b|\cos	heta$,we have $1 \cdot 1 \cdot \cos	heta = \frac{1}{\sqrt{2}}$.$\cos	heta = \frac{1}{\sqrt{2}}$,which implies $	heta = \frac{\pi}{4}$.

If $\theta$ is the angle between the unit vectors $a$ and $b$,then $a - \sqrt{2}b$ will be a unit vector if $\theta = $

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