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