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Question
Show that the points A (−1, 4, −3), B (3, 2, −5), C (−3, 8, −5) and D (−3, 2, 1) are coplanar.
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Solution
\[\text { The points A, B, C and D will be coplanar iff any one of the following triads of vectors are coplanar:} \]
\[ \overrightarrow{AB} , \overrightarrow{AC} , \overrightarrow{AD} ; \overrightarrow{AB} , \overrightarrow{BC} , \overrightarrow{CD} ; \overrightarrow{BC} , \overrightarrow{BA} , \overrightarrow{BD,} \text { etc } . \]
\[\text { To show that }\overrightarrow{AB} , \overrightarrow{AC} , \overrightarrow{AD} \text { are coplanar, we have to prove that their scaler triple product,} \]
\[\text {i . e } . \left[ \overrightarrow{AB} \overrightarrow{AC} \overrightarrow{AD} \right] = 0\]
Now,
\[ \overrightarrow{AB} = \left[ 3 - \left( - 1 \right) \right] \hat {i} + \left( 2 - 4 \right) \hat { j } + \left[ - 5 - \left( - 3 \right) \right] \hat{k} = 4 \hat { i} - 2 \hat {j} - 2 \hat {k} \]
\[ \overrightarrow{AC} = \left[ - 3 - \left( - 1 \right) \right] \hat { i} + \left( 8 - 4 \right) \hat {j} + \left[ - 5 - \left( - 3 \right) \right] \hat{k} = - 2 \hat {i} + 4 \hat {j} - 2 \hat {k} \]
\[ \overrightarrow{AD} = \left[ - 3 - \left( - 1 \right) \right] \hat { i} + \left( 2 - 4 \right) \hat {j} + \left[ 1 - \left( - 3 \right) \right] \hat {k} = - 2\hat { i} - 2 \hat {j} + 4 \hat{k} \]
\[ \therefore \left[ \overrightarrow{AB} \overrightarrow{AC} \overrightarrow{AD} \right] = \begin{vmatrix}4 & - 2 & - 2 \\ - 2 & 4 & - 2 \\ - 2 & - 2 & 4\end{vmatrix} = 4\left( 16 - 4 \right) + 2\left( - 8 - 4 \right) - 2\left( 4 + 8 \right) = 0\]
Thus, the given points are coplanar .
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