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Question
Find the values of 'a' for which the vectors
\[\vec{\alpha} = \hat {i} + 2 \hat {j} + \hat {k} , \vec{\beta} = a \hat {i} + \hat {j} + 2 \hat {k} \text { and } \vec{\gamma} = \hat {i} + 2 \hat {j} + a \hat { k }\] are coplanar.
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Solution
Given:
\[ \vec{\alpha} = \hat {i} + 2 \hat {j} + \hat {k} \]
\[ \vec{\beta} = a \hat {i} + \hat {j} + 2 \hat {k} \]
\[ \vec{\gamma} = \hat {i} + 2 \hat {j} + a \hat {k}\]
\[\text {We know that three vectors } \vec{\alpha} , \vec{\beta} , \vec{\gamma} \text { are coplanar iff their scalar product is zero } . \]
\[ \therefore \left[ \vec{\alpha} \vec{\beta} \vec{\gamma} \right] = 0\]
\[ \Rightarrow \begin{vmatrix}1 & 2 & 1 \\ a & 1 & 2 \\ 1 & 2 & a\end{vmatrix} = 0\]
\[ \Rightarrow 1\left( a - 4 \right) - 2\left( a^2 - 2 \right) + 1\left( 2a - 1 \right) = 0\]
\[ \Rightarrow - 2 a^2 + 3a - 1 = 0\]
\[ \Rightarrow 2 a^2 - 3a + 1 = 0\]
\[ \Rightarrow \left( a - 1 \right)\left( 2a - 1 \right) = 0\]
\[ \Rightarrow a = 1, \frac{1}{2}\]
