Let vec A = (hat i + hat j) and vec B = (2hat | vedclass

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(A) Given $\vec A = \hat i + \hat j$ and $\vec B = 2\hat i - \hat j$.First,calculate the dot product $\vec A \cdot \vec B = (1)(2) + (1)(-1) = 2 - 1 =

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$\sqrt{\frac{5}{9}}$

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(A) Given $\vec A = \hat i + \hat j$ and $\vec B = 2\hat i - \hat j$.First,calculate the dot product $\vec A \cdot \vec B = (1)(2) + (1)(-1) = 2 - 1 = 1$.Let the coplanar vector be $\vec C = a\hat i + b\hat j$.From the condition $\vec A \cdot \vec C = \vec A \cdot \vec B$,we have: $(1)(a) + (1)(b) = 1 \implies a + b = 1 \quad (i)$.From the condition $\vec B \cdot \vec C = \vec A \cdot \vec B$,we have: $(2)(a) + (-1)(b) = 1 \implies 2a - b = 1 \quad (ii)$.Adding equations $(i)$ and $(ii)$: $(a + b) + (2a - b) = 1 + 1 \implies 3a = 2 \implies a = \frac{2}{3}$.Substituting $a = \frac{2}{3}$ into equation $(i)$: $\frac{2}{3} + b = 1 \implies b = 1 - \frac{2}{3} = \frac{1}{3}$.Thus,$\vec C = \frac{2}{3}\hat i + \frac{1}{3}\hat j$.The magnitude of $\vec C$ is $|\vec C| = \sqrt{(\frac{2}{3})^2 + (\frac{1}{3})^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}}$.

Let $\vec A = (\hat i + \hat j)$ and $\vec B = (2\hat i - \hat j)$. The magnitude of a coplanar vector $\vec C$ such that $\vec A \cdot \vec C = \vec B \cdot \vec C = \vec A \cdot \vec B$ is given by

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