Question

Find \[\vec{a} \cdot \vec{b}\] when

\[\vec{a} = \hat{j} + 2 \hat{k}  \text{ and } \vec{b} = 2 \hat{i} + \hat{k}\]

Answer 1

 We have
\[ \vec{a} = \hat{j}+2\hat{k} = 0 \hat{i} + \hat{j} + \hat{2k} \text{ and } \vec{b} =2 \hat{i} + \hat{k} =2\hat {i} +0 \hat{j} + \hat{k}\]
\[ \vec{a} . \vec{b} =\left(\hat {0i} + \hat{j} + \hat{2k}\right).\left( \hat{2i}+\hat{0j} +\hat{k}\right)\]
\[ = \left( 0 \right)\left( 2 \right) + \left( 1 \right)\left( 0 \right) + \left( 2 \right)\left( 1 \right)\]
\[ = 0 + 0 + 2\]
\[ = 2\]