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Question
Show that the vectors \[\vec{a,} \vec{b,} \vec{c}\] given by \[\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} , \vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}\text{ and }\vec{c} = \hat{i} + \hat{j} + \hat{k}\] are non coplanar.
Express vector \[\vec{d} = 2 \hat{i}-j- 3 \hat{k} , \text{ and }\text { as a linear combination of the vectors } \vec{a,} \vec{b}\text{ and }\vec{c} .\]
Solution
Let the given vectors \[\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} , \vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}\] and \[\vec{c} = \hat{i} + {\hat{j} + {\hat{k}}^\dot{}}\] are coplanar. Then one of the vector is expressible as a linear combination of the other two.
Let, \[\hat{i} + 2 \hat{j} + 3 \hat{k} = x \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) + y \left( \hat{i} + \hat{j} + \hat{k} \right) . \]
\[ = \hat{i} \left( 2x + y \right) + \hat{j} \left( x + y \right) + \hat{k} \left( 3x + y \right) .\]
\[\Rightarrow 2x + y = 1, x + y = 2, 3x + y = 3 .\]
On solving the first two equations we get \[x = - 1, y = 3\].
Clearly the values of x, y does not satisfy the third equation.
Hence the given vectors are non-coplanar.
Now,
\[\vec{d} = 2 \hat{i} - \hat{j} - 3 \hat{k}\] which can be expressed as
\[2 \hat{i} - \hat{j} - 3 \hat{k} = x( \hat{i} + 2 \hat{j} + 3 \hat{k} ) + y(2 \hat{i} + \hat{j} + 3 \hat{k} ) + z( \hat{i} + \hat{j} + \hat{k} ) .\]
\[\hat{i} (x + 2y + z) + \hat{j} (2x + y + z) + \hat{k} (3x + 3y + z)\]
\[\Rightarrow x + 2y + z = 2, 2x + y + z = - 1, 3x + 3y + z = - 3 . \]
\[ \Rightarrow x = - \frac{8}{3}, y = \frac{1}{3}, z = 4\]
Hence, \[\vec{d}\] is expressible as the linear combination of \[\vec{a} , \vec{b}\] and \[\vec{c}\]
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