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Question
Show that four points whose position vectors are
\[6 \hat { i} - 7 \hat {j} , 16 \hat { i} - 19 \hat { j} - 4 \hat {k} , 3 \hat {i} - 6 \hat {k} , 2 \hat { i} - 5 \hat {j}+ 10 \hat {k}\]
Solution
DISCLAIMER: Given points are not coplanar.
Let A, B, C, D be the given points . The given points will be coplanar iff any one of
the following triads of vectors are coplanar:
\[ \overrightarrow{AB} , \overrightarrow{AC} , \overrightarrow{AD} ; \overrightarrow{AB} , \overrightarrow{BC} , \overrightarrow{CD} ; \overrightarrow{BC} , \overrightarrow{BA} , \overrightarrow{BD}\text { etc } . \]
\[\text {In order to show that} \overrightarrow{AB} , \overrightarrow{AC} , \overrightarrow{AD} \text {are coplanar, we will have to show that their scaler triple }\]
\[\text {product i . e }. \left[ \overrightarrow{AB} \overrightarrow{AC} \overrightarrow{AD} \right] = 0\]
\[\text {Using}, \vec{PQ} = \text {Position vector of Q - Position vector of P, we obtain }\]
Now,
\[ \overrightarrow{AB} = \left( 16 \hat {i} - 19 \hat {j} - 4 \hat {k} \right) - \left( 6 \hat {i} - 7 \hat {j} \right) = 10 \hat {i} - 12 \hat {j} - 4 \hat {k} \]
\[ \overrightarrow{AC} = \left( 3 \hat {i} - 6 \hat { k} \right) - \left( 6 \hat {i} - 7 \hat {j} \right) = - 3 \hat {i} + 7 \hat {j} - 6 \hat {k} \]
\[\text { and }, \overrightarrow{AD} = \left( 2 \hat {i}- 5 \hat{j} + 10 \hat{k} \right) - \left( 6 \hat {i} - 7 \hat {j} \right) = - 4 \hat{i} + 2 \hat {j} + 10 \hat {k} \]
\[ \therefore \left[ \overrightarrow{AB} \overrightarrow{AC} \overrightarrow{AD} \right] = \begin{vmatrix}10 & - 12 & - 4 \\ - 3 & 7 & - 6 \\ - 4 & 2 & 10\end{vmatrix} = 10\left( 70 + 12 \right) + 12\left( - 30 - 24 \right) - 4\left( - 6 + 28 \right) = 84\]
Thus, the given points are not coplanar .
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