Question

Find λ for which the points A (3, 2, 1), B (4, λ, 5), C (4, 2, −2) and D (6, 5, −1) are coplanar.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{points}A,B,C\mathrm{and}D\mathrm{will}\mathrm{be}\mathrm{coplanar}\mathrm{iff}\mathrm{any}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{traces}\mathrm{of}\mathrm{vectors}\mathrm{are}\mathrm{coplanar}: \\ \overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD};\mathit{}\overrightarrow{AB}\mathit{,}\mathit{}\overrightarrow{BC}\mathit{,}\mathit{}\overrightarrow{CD};\mathit{}\overrightarrow{BC}\mathit{,}\mathit{}\overrightarrow{BA}\mathit{,}\mathit{}\overrightarrow{BD}\mathit{,}\mathrm{etc}\mathit{.} \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\overrightarrow{\mathrm{A}\mathrm{B}},\overrightarrow{\mathrm{A}\mathrm{C}},\overrightarrow{\mathrm{A}\mathrm{D}}\mathrm{are}\mathrm{coplanar}. \\ \mathrm{Thus},\mathrm{their}\mathrm{scaler}\mathrm{triple}\mathrm{product}\left[\overrightarrow{AB}\mathit{}\overrightarrow{AC}\mathit{}\mathit{}\overrightarrow{AD}\right]\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{zero}. \\ \mathrm{Now}, \\ \mathrm{Direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\overrightarrow{PQ}=\left(\mathrm{Direction}\mathrm{ratios}\mathrm{of}\mathrm{vector}Q\right)-\left(\mathrm{Direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{vector}\mathit{}P\right) \\ \mathrm{Direction}\mathrm{ratios}\mathrm{of}\mathrm{vector}\mathit{}\overrightarrow{AB}=\left(4-3\right),\left(\lambda -2\right),\left(5-1\right),\mathrm{i}.\mathrm{e}.1,\mathit{}\lambda -2,4 \\ \mathrm{Direction}\mathrm{ratios}\mathrm{of}\mathrm{vector}\overrightarrow{AC}=\left(4-3\right),\left(2-2\right),\left(-2-1\right),\mathrm{i}.\mathrm{e}.1,0,-3 \\ \mathrm{Direction}\mathrm{ratios}\mathrm{of}\mathrm{vector}\overrightarrow{AD}=\left(6-3\right),\left(5-2\right),\left(-1-1\right),\mathrm{i}.\mathrm{e}.3,3,-2 \\ \therefore \left[\overrightarrow{\mathrm{A}\mathrm{B}}\mathit{}\overrightarrow{\mathrm{A}\mathrm{C}}\mathit{}\mathit{}\overrightarrow{\mathrm{A}\mathrm{D}}\right]=\left|\begin{array}{ccc}1& \lambda -2& 4\\ 1& 0& -3\\ 3& 3& -2\end{array}\right|=1\left[0-\left(-9\right)\right]-\left(\lambda -2\right)\left[-2-\left(-9\right)\right]+4\left(3-0\right)=0 \\ \Rightarrow 7\lambda =35 \\ \Rightarrow \lambda =5 \end{gathered}$$