Let p=hat{i}+2hat{j}-hat{k} and q=2hat{i}-hat | vedclass

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(A) Given equations are $p=a-2b$ and $q=2a+b$.To solve for $a$, multiply the second equation by $2$: $2q=4a+2b$.Adding this to the first equation: $p+

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$\cos^{-1}\left(\frac{3}{2\sqrt{221}}\right)$

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(A) Given equations are $p=a-2b$ and $q=2a+b$.To solve for $a$, multiply the second equation by $2$: $2q=4a+2b$.Adding this to the first equation: $p+2q = (a-2b) + (4a+2b) = 5a$.$5a = (\hat{i}+2\hat{j}-\hat{k}) + 2(2\hat{i}-\hat{j}+\hat{k}) = 5\hat{i}+\hat{k}$.Thus, $a = \hat{i} + \frac{1}{5}\hat{k}$.To solve for $b$, substitute $a$ into $q=2a+b$: $b = q-2a$.$b = (2\hat{i}-\hat{j}+\hat{k}) - 2(\hat{i} + \frac{1}{5}\hat{k}) = -\hat{j} + \frac{3}{5}\hat{k}$.The dot product $a \cdot b = (1)(0) + (0)(-1) + (\frac{1}{5})(\frac{3}{5}) = \frac{3}{25}$.The magnitudes are $|a| = \sqrt{1^2 + (\frac{1}{5})^2} = \sqrt{\frac{26}{25}} = \frac{\sqrt{26}}{5}$ and $|b| = \sqrt{(-1)^2 + (\frac{3}{5})^2} = \sqrt{\frac{34}{25}} = \frac{\sqrt{34}}{5}$.The angle $	heta$ is given by $\cos 	heta = \frac{a \cdot b}{|a||b|} = \frac{3/25}{(\sqrt{26}/5)(\sqrt{34}/5)} = \frac{3}{\sqrt{26 	imes 34}} = \frac{3}{\sqrt{884}} = \frac{3}{2\sqrt{221}}$.Therefore, $	heta = \cos^{-1}\left(\frac{3}{2\sqrt{221}}ight)$.

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

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