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Question
Show of the following triad of vector is coplanar:
\[\vec{a} = - 4 \hat{i} - 6 \hat{j} - 2 \hat{k} , \vec{b} = -\hat{ i} + 4 \hat{j} + 3 \hat{k} , \vec{c} = - 8 \hat{i} - \hat{j} + 3 \hat{k}\]
Solution
Given:
\[ \vec{a} = - 4 \hat{i}- 6 \hat {j} - 2 \hat {k} \]
\[ \vec{b} = -\text { i } + 4\hat { j} + 3 \hat {k}\]
\[ \vec{c} = - 8 \hat {i} - \hat{j} + 3\hat { k} \]
\[\text { We know that three vectors} \vec{a} , \vec{b} , \vec{c} \text {are coplanar iff their scalar triple product is zero, i . e} . \left[ \vec{a} \vec{b} \vec{c} \right] = 0 . \]
Here,
\[\left[ \vec{a} \vec{b} \vec{c} \right] = \begin{vmatrix}- 4 & - 6 & - 2 \\ - 1 & 4 & 3 \\ - 8 & - 1 & 3\end{vmatrix} = - 4\left( 12 + 3 \right) + 6\left( - 3 + 24 \right) - 2\left( 1 + 32 \right) = 0\]
Hence, the given vectors are coplanar .
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