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.

Answer 1

Given:

\[ \vec{\alpha} = \hat {i} + 2 \hat {j} + \hat {k} \]

\[ \vec{\beta} = a \hat {i} + \hat {j} + 2 \hat {k} \]

\[ \vec{\gamma} = \hat {i} + 2 \hat {j} + a \hat {k}\]

\[\text {We know that three vectors } \vec{\alpha} , \vec{\beta} , \vec{\gamma} \text { are coplanar iff their scalar product is zero } . \]

\[ \therefore \left[ \vec{\alpha} \vec{\beta} \vec{\gamma} \right] = 0\]

\[ \Rightarrow \begin{vmatrix}1 & 2 & 1 \\ a & 1 & 2 \\ 1 & 2 & a\end{vmatrix} = 0\]

\[ \Rightarrow 1\left( a - 4 \right) - 2\left( a^2 - 2 \right) + 1\left( 2a - 1 \right) = 0\]

\[ \Rightarrow - 2 a^2 + 3a - 1 = 0\]

\[ \Rightarrow 2 a^2 - 3a + 1 = 0\]

\[ \Rightarrow \left( a - 1 \right)\left( 2a - 1 \right) = 0\]

\[ \Rightarrow a = 1, \frac{1}{2}\]