Question

If \[\vec{\alpha} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \text{ and } \vec{\beta} = 2 \hat{i} + \hat{j} - 4 \hat{k} ,\] then express \[\vec{\beta}\] in the form of  \[\vec{\beta} = \vec{\beta_1} + \vec{\beta_2} ,\]  where \[\vec{\beta_1}\] is parallel to \[\vec{\alpha} \text{ and } \vec{\beta_2}\]  is perpendicular to \[\vec{\alpha}\]

Answer 1

\[\text{ Given that } \vec{\alpha} =3 \hat{i} + 4 \hat{j} +5 \hat{k} \text{ and } \vec{\beta} =2 \hat{i} + \hat{j} - 4 \hat{k} \]
\[\hat{ Also },\]
\[ \vec{\beta} = \vec{\beta_1} + \vec{\beta_2} \]
\[ \Rightarrow \vec{\beta_2} = \vec{\beta} - \vec{\beta}_1 . . . \left( 1 \right)\]
\[\text{ Since } \vec{\beta}_1 \text{ is parallel to } \vec{\alpha} ,\]
\[ \vec{\beta_1} = t \vec{\alpha} \]
\[ \Rightarrow \vec{\beta_1} = t \left( 3 \hat{i} + 4 \hat{j} +5 \hat{k} \right) = 3t \hat{i} + 4t \hat{j} +5t \hat{k} ...(2)\]
\[\text{ Substituting the values of } \vec{\beta_1} \text{ and } \vec{\alpha} \text{ in } (1), \text{ we get }\]
\[ \vec{\beta_2} = 2 \hat{i} + \hat{j} - 4 \hat{k} - \left( 3t \hat{i} + 4t \hat{j} +5t \hat{k} \right) = \left( 2 - 3t \right) \hat{i} + \left( 1 - 4t \right) \hat{j} + \left( - 4 - 5t \right) \hat{k} . . . \left( 3 \right)\]
\[\text{ Since } \vec{\beta_2} \text{ is perpendicular to } \vec{\alpha} ,\]
\[ \vec{\beta_2} . \vec{\alpha} = 0\]
\[ \Rightarrow \left[ \left( 2 - 3t \right) \text{i} + \left( 1 - 4t \right) \hat{j} + \left( - 4 - 5t \right) \hat{k} \right] . \left( 3 \hat{i} + 4 \hat{j} +5 \hat{k} \right) = 0\]
\[ \Rightarrow 3 \left( 2 - 3t \right) + 4 \left( 1 - 4t \right) + 5 \left( - 4 - 5t \right) = 0\]
\[ \Rightarrow 6 - 9t + 4 - 16t - 20 - 25t = 0\]
\[ \Rightarrow - 50t = 10\]
\[ \Rightarrow t = \frac{- 1}{5}\]
\[\text{ From } (2) \text{ and } (3), \text{ we get }\]
\[ \vec{\beta_1} = \frac{- 1}{5} \left( 3 \hat{i} + 4 \hat{j} +5 \hat{k} \right)\]
\[ \vec{\beta_2} = \frac{13}{5} \hat{i} + \frac{9}{5} \hat{j} - 3 \hat{k} = \frac{1}{5}\left( 13 \hat{i} + 9 \hat{j} - 15 \hat{k} \right)\]