Question

Find the value of λ so that the following vector is coplanar:

\[\vec{a} = \hat{i} + 2\hat { j} - 3 \hat {k} , \vec{b} = 3 \hat{i} + \lambda \hat {j} + \hat {k} , \vec{c} = \hat {i} + 2 \hat {j} + 2 \hat {k}\]

Answer 1

Given: 

\[ \vec{a} = \hat{i} + 2 \hat {j} - 3 \hat{k} \]

\[ \vec{b} = 3 \hat {i} + \lambda \hat { j} + \hat {k} \]

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

\[\text {We know that vectors} \vec{a} , \vec{b} , \vec{c} \text { are coplanar iff} \left[ \vec{a} \vec{b} \vec{c} \right] = 0 . \]

\[\text { It is given that } \vec{a} , \vec{b} , \vec{c} \text { are coplanar } . \]

\[ \therefore \left[ \vec{a} \vec{b} \vec{c} \right] = 0\]

\[ \Rightarrow \begin{vmatrix}1 & 2 & - 3 \\ 3 & \lambda & 1 \\ 1 & 2 & 2\end{vmatrix} = 0 \]

\[ \Rightarrow 1\left( 2\lambda - 2 \right) - 2\left( 6 - 1 \right) - 3\left( 6 - \lambda \right) = 0\]

\[ \Rightarrow 5\lambda - 30 = 0 \]

\[ \Rightarrow \lambda = 6\]