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Scalar Triple Product — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSScalar Triple Product5 Marks
Question
If four points A, B, C and D with position vectors $4\hat{\text{i}}+3\hat{\text{j}},5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}},5\hat{\text{i}}+3\hat{\text{j}}$ and $7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}}$ respectively are coplanar, then find the value of x.
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Answer

Let $\vec{\text{OA}}=4\vec{\text{i}}+3\vec{\text{j}}+3\vec{\text{k}},\vec{\text{OB}}=5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}},\vec{\text{OC}}=5\hat{\text{i}}+3\hat{\text{j}}$ and $7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}}.$
$\therefore\vec{\text{AB}}=(5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}})-(4\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})\\=\hat{\text{i}}+(\text{x}-3)\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{AC}}=(5\hat{\text{i}}+3\hat{\text{j}})-(4\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})\\=\hat{\text{i}}-3\hat{\text{k}}$
$\vec{\text{AD}}=(7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}})-(4\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})\\=3\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$
Since the given four points are coplanar, so the vectors $\vec{\text{AB}},\vec{\text{AC}}$ and $\vec{\text{AD}}$ are also coplanar.
$\therefore\big[\vec{\text{AB}}\vec{\text{ AC }}\vec{\text{AD}}\big]=0$
$\begin{vmatrix}1&\text{x}-3&4\\1&0&-3\\3&3&-2 \end{vmatrix}=0$
$\Rightarrow 1(0+9)-(\text{x}-3)(-2+9)+4(3-0)=0$
$\Rightarrow 9-7\text{x}+21+12=0$
$\Rightarrow7\text{x}=42$
$\Rightarrow \text{x}=6$
Thus, the value of x is 6.

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