Two unit vectors hat{a}_{1} and hat{a}_{2} ar | vedclass

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(D) Given that $\hat{a}_{1}$ and $\hat{a}_{2}$ are unit vectors,so $|\hat{a}_{1}| = 1$ and $|\hat{a}_{2}| = 1$.Given $|\hat{a}_{1}-\hat{a}_{2}| = \sqr

Vedclass · Accepted Answer

$4.5$

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(D) Given that $\hat{a}_{1}$ and $\hat{a}_{2}$ are unit vectors,so $|\hat{a}_{1}| = 1$ and $|\hat{a}_{2}| = 1$.Given $|\hat{a}_{1}-\hat{a}_{2}| = \sqrt{3}$.Squaring both sides,we get $|\hat{a}_{1}-\hat{a}_{2}|^2 = 3$.$(\hat{a}_{1}-\hat{a}_{2}) \cdot (\hat{a}_{1}-\hat{a}_{2}) = 3 \implies |\hat{a}_{1}|^2 + |\hat{a}_{2}|^2 - 2(\hat{a}_{1} \cdot \hat{a}_{2}) = 3$.$1 + 1 - 2(\hat{a}_{1} \cdot \hat{a}_{2}) = 3 \implies 2 - 2(\hat{a}_{1} \cdot \hat{a}_{2}) = 3 \implies \hat{a}_{1} \cdot \hat{a}_{2} = -1/2$.Now,we need to evaluate $(\hat{a}_{1}-\hat{a}_{2}) \cdot (2\hat{a}_{1}-\hat{a}_{2})$.$= 2(\hat{a}_{1} \cdot \hat{a}_{1}) - (\hat{a}_{1} \cdot \hat{a}_{2}) - 2(\hat{a}_{2} \cdot \hat{a}_{1}) + (\hat{a}_{2} \cdot \hat{a}_{2})$.$= 2(1) - 3(\hat{a}_{1} \cdot \hat{a}_{2}) + 1$.$= 3 - 3(-1/2) = 3 + 3/2 = 9/2 = 4.5$.

Column $I$	Column $II$
$(A) (A+B)$	$(p)$ North-east
$(B) (A-B)$	$(q)$ Vertically upwards
$(C) (A \times B)$	$(r)$ Vertically downwards
$(D) (A \times B) \times (A \times B)$	$(s)$ None

Two unit vectors $\hat{a}_{1}$ and $\hat{a}_{2}$ are inclined to each other at an angle $\theta$. If $|\hat{a}_{1}-\hat{a}_{2}|=\sqrt{3}$,then the value of $(\hat{a}_{1}-\hat{a}_{2}) \cdot (2\hat{a}_{1}-\hat{a}_{2})$ is:

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