Question

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

 

Answer 1

\[\text{ Let } \vec{a} =2 \hat{i} - \hat{j} + 3 \hat{k} \text{ and } \vec{b} =2 \hat{i} + 4 \hat{j} - 2 \hat{k} \]
\[\text{ and } \vec{x} \text{ and } \vec{y} \text{ be such that }\]
\[ \vec{a} = \vec{x} + \vec{y} \]
\[ \Rightarrow \vec{y} = \vec{a} - \vec{x} ...(1)\]
\[\text{ Since } \vec{x} \text{ is parallel to } \vec{b} ,\]
\[ \vec{x} = t \vec{b} \]
\[ \Rightarrow \vec{x} = t \left( 2 \text{i} + 4\hat{j} - 2 \hat{k} \right) = 2t \hat{i} + 4t \hat{j} - 2t \hat{k} ..............(2)\]

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