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Scalar Triple Product — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSScalar Triple Product5 Marks
Question
Show that four points whose position vectors are
$6\hat{\text{i}}-7\hat{\text{j}},16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},3\hat{\text{i}}-6\hat{\text{k}},2\hat{\text{i}}-5\hat{\text{j}}+10\hat{\text{k}}$ are coplanar.
✓

Answer

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

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