Express $2 \hat{i} - \hat{j} + 3 \hat{k}$ as the sum of a vector parallel and a vector perpendicular to $2 \hat{i} + 4 \hat{j} - 2 \hat{k} .$

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Solution

$Let \vec{a} =2 \hat{i} - \hat{j} + 3 \hat{k} and \vec{b} = 2 \hat{i} + 4 \hat{j} - 2 \hat{k} and \vec{x} and \vec{y} be such that \vec{a} = \vec{x} + \vec{y} \Rightarrow \vec{y} = \vec{a} - \vec{x} ...(1) Since \vec{x} is parallel to \vec{b}, \vec{x} = t \vec{b} \Rightarrow \vec{x} = t (2 \hat{i} + 4 \hat{j} - 2 \hat{k}) = 2 t \hat{i} + 4 t \hat{j} - 2 t \hat{k} ...(2) Substituting the values of \vec{x} and \vec{a} in (1), \vec{y} = 2 \hat{i} - \hat{j} + 3 \hat{k} - (2 t \hat{i} + 4 t \hat{j} - 2 t \hat{k}) = (2 - 2 t) \hat{i} + (- 1 - 4 t) \hat{j} + (3 + 2 t) \hat{k} . . . (3) Since \vec{y} is perpendicular to \vec{b}, \vec{y} . \vec{b} = 0 \Rightarrow [(2 - 2 t) \hat{i} + (- 1 - 4 t) \hat{j} + (3 + 2 t) \hat{k}] . (2 \hat{i} + 4 \hat{j} - 2 \hat{k}) = 0 \Rightarrow 2 (2 - 2 t) + 4 (- 1 - 4 t) - 2 (3 + 2 t) = 0 \Rightarrow 4 - 4 t - 4 - 16 t - 6 - 4 t = 0 \Rightarrow - 24 t = 6 \Rightarrow t = (\frac{- 1}{4}) From (2) and (3), \vec{x} = 2 (\frac{- 1}{4}) \hat{i} + 4 (\frac{- 1}{4}) \hat{j} - 2 (\frac{- 1}{4}) \hat{k} = \frac{- 1}{2} \hat{i} - \hat{j} + \frac{1}{2} \hat{k} \vec{y} = [2 - 2 (\frac{- 1}{4})] \hat{i} + [- 1 - 4 (\frac{- 1}{4})] \hat{j} + [3 + 2 (\frac{- 1}{4})] \hat{k} = \frac{5}{2} \hat{i} + \frac{5}{2} \hat{k} = \frac{5}{2} (\hat{i} + \hat{k}) So, \vec{a} = \vec{x} + \vec{y} = (\frac{- 1}{2} \hat{i} - \hat{j} + \frac{1}{2} \hat{k}) + \frac{5}{2} (\hat{i} + \hat{k})$

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