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Polynomials

If α→=3 i ∧+4...

Question

If $\vec{α} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} and \vec{β} = 2 \hat{i} + \hat{j} - 4 \hat{k},$ then express $\vec{β}$ in the form of $\vec{β} = \vec{β_{1}} + \vec{β_{2}},$ where $\vec{β_{1}}$ is parallel to $\vec{α} and \vec{β_{2}}$ is perpendicular to $\vec{α}$ .

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Solution

$Given that \vec{α} = 3 \hat{i} + 4 \hat{j} +5 \hat{k} and \vec{β} =2 \hat{i} + \hat{j} - 4 \hat{k} Also, \vec{β} = \vec{β_{1}} + \vec{β_{2}} \Rightarrow \vec{β_{2}} = \vec{β} - {\vec{β}}_{1} . . . (1) Since {\vec{β}}_{1} is parallel to \vec{α}, \vec{β_{1}} = t \vec{α} \Rightarrow \vec{β_{1}} = t (3 \hat{i} + 4 \hat{j} +5 \hat{k}) = 3 t \hat{i} + 4 t \hat{j} +5 t \hat{k} ...(2) Substituting the values of \vec{β_{1}} and \vec{α} in (1), we get \vec{β_{2}} = 2 \hat{i} + \hat{j} - 4 \hat{k} - (3 t \hat{i} + 4 t \hat{j} +5 t \hat{k}) = (2 - 3 t) \hat{i} + (1 - 4 t) \hat{j} + (- 4 - 5 t) \hat{k} . . . (3) Since \vec{β_{2}} is perpendicular to \vec{α}, \vec{β_{2}} . \vec{α} = 0 \Rightarrow [(2 - 3 t) \hat{i} + (1 - 4 t) \hat{j} + (- 4 - 5 t) \hat{k}] . (3 \hat{i} + 4 \hat{j} +5 \hat{k}) = 0 \Rightarrow 3 (2 - 3 t) + 4 (1 - 4 t) + 5 (- 4 - 5 t) = 0 \Rightarrow 6 - 9 t + 4 - 16 t - 20 - 25 t = 0 \Rightarrow - 50 t = 10 \Rightarrow t = \frac{- 1}{5} From (2) and (3), we get \vec{β_{1}} = \frac{- 1}{5} (3 \hat{i} + 4 \hat{j} +5 \hat{k}) \vec{β_{2}} = \frac{13}{5} \hat{i} + \frac{9}{5} \hat{j} - 3 \hat{k} = \frac{1}{5} (13 \hat{i} + 9 \hat{j} - 15 \hat{k})$

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