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Scalar Triple Product — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSScalar Triple Product3 Marks
Question
Show that the four points having position vectors
$6\hat{\text{i}}-7\hat{\text{j}},16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},3\hat{\text{j}}-6\hat{\text{k}},2\hat{\text{i}}+5\hat{\text{j}}+10\hat{\text{k}}$ are not coplanar.
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Answer

Let
$\text{OA}=6\hat{\text{i}}-7\hat{\text{j}},{\text{OB}}=16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},$
$\text{OC}=3\hat{\text{j}}-6\hat{\text{k}},\text{OD}=2\hat{\text{i}}+5\hat{\text{j}}+10\hat{\text{k}}$
$\text{AB}=\text{OB}-\text{OA}=16\hat{\text{i}}-25\hat{\text{j}}-4\hat{\text{k}}$
$\text{AC}=\text{OC}-\text{OA}=-16\hat{\text{i}}-16\hat{\text{j}}+2\hat{\text{k}}$
$\text{CD}=\text{OD}-\text{OC}=2\hat{\text{i}}+2\hat{\text{j}}+16\hat{\text{k}}$
$\text{AD}=\text{OD}-\text{OA}=4\hat{\text{i}}+12\hat{\text{j}}+10\hat{\text{k}}$
The four points are co-planer if vectors $\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}}$ are co-planer.
$\begin{vmatrix}16&-25&-4\\-16&-16&2\\-4&12&10 \end{vmatrix}$
$=16(-160-24)+25(-160+8)-4(-144+64)$
$\neq 0$
Hence the point are not co-planar.

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