If vec{p} = hat{i} + hat{j} + hat{k} and vec{ | vedclass

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(C) Given equations are:$2\vec{a} + \vec{b} = \hat{i} + \hat{j} + \hat{k} \quad \dots(1)$$\vec{a} + 2\vec{b} = \hat{i} + \hat{j} - \hat{k} \quad \dots

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$\cos^{-1}\left(-\frac{7}{11}\right)$

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(C) Given equations are:$2\vec{a} + \vec{b} = \hat{i} + \hat{j} + \hat{k} \quad \dots(1)$$\vec{a} + 2\vec{b} = \hat{i} + \hat{j} - \hat{k} \quad \dots(2)$Multiplying $(1)$ by $2$,we get $4\vec{a} + 2\vec{b} = 2\hat{i} + 2\hat{j} + 2\hat{k}$.Subtracting $(2)$ from this,we get $3\vec{a} = \hat{i} + \hat{j} + 3\hat{k} \implies \vec{a} = \frac{1}{3}(\hat{i} + \hat{j} + 3\hat{k})$.Substituting $\vec{a}$ in $(1)$,$\vec{b} = (\hat{i} + \hat{j} + \hat{k}) - 2\vec{a} = (\hat{i} + \hat{j} + \hat{k}) - \frac{2}{3}(\hat{i} + \hat{j} + 3\hat{k}) = \frac{1}{3}(\hat{i} + \hat{j} - 3\hat{k})$.The angle $	heta$ between $\vec{a}$ and $\vec{b}$ is given by $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.$\vec{a} \cdot \vec{b} = \frac{1}{9}(1 + 1 - 9) = -\frac{7}{9}$.$|\vec{a}| = \frac{1}{3}\sqrt{1^2 + 1^2 + 3^2} = \frac{\sqrt{11}}{3}$ and $|\vec{b}| = \frac{1}{3}\sqrt{1^2 + 1^2 + (-3)^2} = \frac{\sqrt{11}}{3}$.$\cos 	heta = \frac{-7/9}{(\sqrt{11}/3)(\sqrt{11}/3)} = \frac{-7/9}{11/9} = -\frac{7}{11}$.Therefore,$	heta = \cos^{-1}\left(-\frac{7}{11}ight)$.

If $\vec{p} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{q} = \hat{i} + \hat{j} - \hat{k}$,and $\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{p} = 2\vec{a} + \vec{b}$ and $\vec{q} = \vec{a} + 2\vec{b}$,then the angle between $\vec{a}$ and $\vec{b}$ is:

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