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Question
Find \[\left[ \vec{a} \vec{b} \vec{c} \right]\] , when \[\vec{a} = 2 \hat{i} - 3 \hat{j} , \vec{b} = \hat{i} + \hat{j} - \hat{k} \text{ and } \vec{c} = 3 \hat{i} - \hat{k}\]
Solution
Given :
\[ \vec{a} = 2 \hat{ i } - 3 \hat{j} \]
\[ \vec{b} =\hat{ i} + \hat{j} - \hat{k} \]
\[ \vec{c} = 3 \hat{i}- \hat{k} \]
\[ \therefore \vec{a} \times \vec{b} = \left( 2\hat{ i} - 3 \hat{j} \right) \times \left( \hat{i} + \hat{j} - \hat{k} \right)\]
\[ = 2 \hat{k} + 2 \hat{j} + 3 \hat{k} + 3 \hat{i} \]
\[ = 3 \hat{i} + 2 \hat{j} + 5 \hat{k} \]
\[\left( \vec{a} \times \vec{b} \right) . \vec{c} = \left( 3\hat{ i} + 2 \hat{j} + 5\hat{ k} \right) . \left( 3 \hat{i} - \hat{k} \right)\]
\[ = 9 - 5 = 4 . . . \left( 1 \right)\]
Now,
\[\left[ \vec{a} \vec{b} \vec{c} \right] = \left( \vec{a} \times \vec{b} \right) . \vec{c} \]
\[ = 4 \left[ \text{Using} \left( 1 \right) \right]\]
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