Question

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

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

Answer 1

\[\left( i \right) Given: \]

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

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

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

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

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

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

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

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

\[ \Rightarrow \lambda - 1 + 3\lambda - 2 - \lambda = 0\]

\[ \Rightarrow 3\lambda - 3 = 0 \]

\[ \Rightarrow \lambda = 1\]