Given vec{a}=3 hat{i}-hat{j},vec{b}=2 hat{i}+ | vedclass

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(A) We are given $\vec{a} = 3 \hat{i} - \hat{j}$ and $\vec{b} = 2 \hat{i} + \hat{j} - 3 \hat{k}$.Since $\overrightarrow{b_1}$ is parallel to $\vec{a}$

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$\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k}$

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(A) We are given $\vec{a} = 3 \hat{i} - \hat{j}$ and $\vec{b} = 2 \hat{i} + \hat{j} - 3 \hat{k}$.Since $\overrightarrow{b_1}$ is parallel to $\vec{a}$,we can write $\overrightarrow{b_1} = \lambda \vec{a} = \lambda(3 \hat{i} - \hat{j}) = 3\lambda \hat{i} - \lambda \hat{j}$.We know $\vec{b} = \overrightarrow{b_1} + \overrightarrow{b_2}$,so $\overrightarrow{b_2} = \vec{b} - \overrightarrow{b_1} = (2 - 3\lambda) \hat{i} + (1 + \lambda) \hat{j} - 3 \hat{k}$.Since $\overrightarrow{b_2}$ is perpendicular to $\vec{a}$,their dot product is zero: $\overrightarrow{b_2} \cdot \vec{a} = 0$.$(2 - 3\lambda)(3) + (1 + \lambda)(-1) + (-3)(0) = 0$.$6 - 9\lambda - 1 - \lambda = 0$.$5 - 10\lambda = 0 \Rightarrow \lambda = \frac{1}{2}$.Substituting $\lambda = \frac{1}{2}$ into the expression for $\overrightarrow{b_2}$:$\overrightarrow{b_2} = (2 - 3(\frac{1}{2})) \hat{i} + (1 + \frac{1}{2}) \hat{j} - 3 \hat{k}$.$\overrightarrow{b_2} = (2 - \frac{3}{2}) \hat{i} + (\frac{3}{2}) \hat{j} - 3 \hat{k} = \frac{1}{2} \hat{i} + \frac{3}{2} \hat{j} - 3 \hat{k}$.

Given $\vec{a}=3 \hat{i}-\hat{j}$,$\vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}$ and $\vec{b}=\overrightarrow{b_1}+\overrightarrow{b_2}$,where $\overrightarrow{b_1}$ is parallel to $\vec{a}$ and $\overrightarrow{b_2}$ is perpendicular to $\vec{a}$,then $\overrightarrow{b_2}$ is equal to

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