Question

If \[\vec{a} = \hat{i} + \hat{j} + \hat{k} , \vec{b} = 4 \hat{i} - 2 \hat{j} + 3 \hat{k} \text { and } \vec{c} = \hat{i} - 2 \hat{j} + \hat{k} ,\] find a vector of magnitude 6 units which is parallel to the vector \[2 \vec{a} - \vec{b} + 3 \vec{c .}\]

Answer 1

We have, \[\vec{a} = \hat{i} + \hat{j} + \hat{k} , \vec{b} = 4 \hat{i} - 2 \hat{j} + 3 \hat{k}\] and \[\vec{c} = \hat{i} - 2 \hat{j} + \hat{k} .\] 
Then, 
\[2 \vec{a} - \vec{b} + 3 \vec{c} = 2\left( \hat{i} + \hat{j} + \hat{k} \right) - \left( 4 \hat{i} - 2 \hat{j} + 3 \hat{k} \right) + 3\left( \hat{i} - 2 \hat{j} + \hat{k} \right) = \hat{i} - 2 \hat{j} + 2 \hat{k}. \]
∴ A unit vector parallel to \[2 \vec{a} - \vec{b} + 3 \vec{c}\] is \[\frac{2 \vec{a} - \vec{b} + 3 \vec{c}}{\left| 2 \vec{a} - \vec{b} + 3 \vec{c} \right|}\]
\[= \frac{\left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right)}{\sqrt{1^2 + \left( - 2 \right)^2 + 2^2}}\]
\[= \frac{\left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right)}{\sqrt{9}} \]
\[ = \frac{\left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right)}{3}\]
Hence, Required vector = \[\frac{6}{3}\left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) = 2 \hat{i} - 4 \hat{j} + 4 \hat{k} .\]