Given that a and b are two unit non-collinear | vedclass

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(D) Let $	heta$ be the angle between $a$ and $b$. Since $a$ and $b$ are unit vectors,$|a| = 1$ and $|b| = 1$.$v = a 	imes b$,so $|v| = |a||b| \sin \

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Both $(A)$ and $(C)$

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(D) Let $	heta$ be the angle between $a$ and $b$. Since $a$ and $b$ are unit vectors,$|a| = 1$ and $|b| = 1$.$v = a 	imes b$,so $|v| = |a||b| \sin 	heta = \sin 	heta$.Now,$u = a - (a \cdot b)b = a - (\cos 	heta)b$.Calculating $|u|^2 = u \cdot u = (a - \cos 	heta \, b) \cdot (a - \cos 	heta \, b) = |a|^2 + \cos^2 	heta |b|^2 - 2 \cos 	heta (a \cdot b) = 1 + \cos^2 	heta - 2 \cos^2 	heta = 1 - \cos^2 	heta = \sin^2 	heta$.Thus,$|u| = \sin 	heta$.Comparing $|v|$ and $|u|$,we get $|v| = |u|$.Also,$u \cdot b = (a - \cos 	heta \, b) \cdot b = a \cdot b - \cos 	heta (b \cdot b) = \cos 	heta - \cos 	heta = 0$.Therefore,$|u| + |u \cdot b| = |u| + 0 = |u| = |v|$.Both options $(A)$ and $(C)$ are correct.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$,then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then,its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$,then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

Given that $a$ and $b$ are two unit non-collinear vectors,if $u = a - (a \cdot b)b$ and $v = a \times b$,then find $|v| =$.

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