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Question
Show that the four points having position vectors
\[6 \hat { i} - 7 \hat { j} , 16 \hat {i} - 19 \hat {j}- 4 \hat {k} , 3 \hat {j} - 6 \hat {k} , 2 \hat {i} + 5 \hat {j} + 10 \hat {k}\] are not coplanar.
Solution
Let A, B, C and D be the given points . These points will be coplanar iff any one of the following triads of vectors are coplanar:
\[ \overrightarrow{AB} , \overrightarrow{AC} , \overrightarrow{AD} ; \overrightarrow{AB} , \overrightarrow{BC} , \overrightarrow{CD} ; \overrightarrow{BC} , \overrightarrow{BA} , \overrightarrow{BD,} etc . \]
\[\text { To show that } \overrightarrow{AB} , \overrightarrow{AC} , \overrightarrow{AD}\text { are not coplanar, we have to prove that their scaler triple product }, \]
\[i . e . \left[ \overrightarrow{AB} \overrightarrow{AC} \overrightarrow{AD} \right] \neq 0\]
Now,
\[ \overrightarrow{PQ} = \left( \text { Position vector of Q } \right) - \left( \text { Position vector of P } \right)\]
\[ \overrightarrow{AB} = \left( 16 \hat { i}- 19 \hat { j} - 4 \hat { k} \right) - \left( 6 \hat {i} - 7 \hat {j} \right) = 10 \hat {i} - 12 \hat{j} - 4 \hat { k} \]
\[ \overrightarrow{AC} = \left( 3 \hat {j} - 6 \hat { k} \right) - \left( 6 \hat {i} - 7 \hat {j} \right) = - 6 \hat {i} + 10\hat { j} - 6 \hat {k} \]
\[ \overrightarrow{AD} = \left( 2 \hat { i} + 5\hat { j} + 10 \hat{k} \right) - \left( 6 \hat { i} - 7 \hat {j} \right) = - 4 \hat {i} + 12 \hat {j} + 10 \hat {k} \]
\[ \therefore \left[ \overrightarrow{AB} \overrightarrow{AC} \overrightarrow{AD} \right] = \begin{vmatrix}10 & - 12 & - 4 \\ - 6 & 10 & - 6 \\ - 4 & 12 & 10\end{vmatrix} = 10\left( 100 + 72 \right) + 12\left( - 60 - 24 \right) - 4\left( - 72 + 40 \right) = 840 \neq 0\]
Thus, the given points are not coplanar .
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