If a=2 hat{i}+hat{j}-3 hat{k} and b=3 hat{i}- | vedclass

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(C) Given vectors are $a = 2\hat{i} + \hat{j} - 3\hat{k}$ and $b = 3\hat{i} - \hat{j} + 2\hat{k}$.First,calculate $2a + b$:$2a + b = 2(2\hat{i} + \hat

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$\cos ^{-1}\left(\frac{51}{66}\right)$

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(C) Given vectors are $a = 2\hat{i} + \hat{j} - 3\hat{k}$ and $b = 3\hat{i} - \hat{j} + 2\hat{k}$.First,calculate $2a + b$:$2a + b = 2(2\hat{i} + \hat{j} - 3\hat{k}) + (3\hat{i} - \hat{j} + 2\hat{k}) = (4\hat{i} + 2\hat{j} - 6\hat{k}) + (3\hat{i} - \hat{j} + 2\hat{k}) = 7\hat{i} + \hat{j} - 4\hat{k}$.Next,calculate $a + 2b$:$a + 2b = (2\hat{i} + \hat{j} - 3\hat{k}) + 2(3\hat{i} - \hat{j} + 2\hat{k}) = (2\hat{i} + \hat{j} - 3\hat{k}) + (6\hat{i} - 2\hat{j} + 4\hat{k}) = 8\hat{i} - \hat{j} + \hat{k}$.Let $u = 2a + b = 7\hat{i} + \hat{j} - 4\hat{k}$ and $v = a + 2b = 8\hat{i} - \hat{j} + \hat{k}$.The angle $	heta$ between $u$ and $v$ is given by $\cos 	heta = \frac{u \cdot v}{|u| |v|}$.$u \cdot v = (7)(8) + (1)(-1) + (-4)(1) = 56 - 1 - 4 = 51$.$|u| = \sqrt{7^2 + 1^2 + (-4)^2} = \sqrt{49 + 1 + 16} = \sqrt{66}$.$|v| = \sqrt{8^2 + (-1)^2 + 1^2} = \sqrt{64 + 1 + 1} = \sqrt{66}$.Therefore,$\cos 	heta = \frac{51}{\sqrt{66} \sqrt{66}} = \frac{51}{66}$.Thus,$	heta = \cos^{-1}\left(\frac{51}{66}ight)$.

If $a=2 \hat{i}+\hat{j}-3 \hat{k}$ and $b=3 \hat{i}-\hat{j}+2 \hat{k}$,then find the angle between the vectors $2 a+b$ and $a+2 b$.

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