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Scalar Triple Product — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSScalar Triple Product5 Marks
Question
Find the value of $\lambda$ for which the four points with position vectors
$-\hat{\text{j}}-\hat{\text{k}},4\hat{\text{i}}+5\hat{\text{j}}+\lambda\hat{\text{k}},3\hat{\text{i}}+9\hat{\text{j}}+4\hat{\text{k}}$ and $-4\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$ are co planar.
✓

Answer

Let
position vector of $\text{A}=-\hat{\text{j}}-\hat{\text{k}}$
position vector of $\text{B}=4\hat{\text{i}}+5\hat{\text{j}}+\lambda\hat{\text{k}}$
position vector of $\text{C}=3\hat{\text{i}}+9\hat{\text{j}}+4\hat{\text{k}}$
position vector of $\text{D}=-4\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$
The four points are coplanar if the vectors $\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}}$ are coplanar.
$\vec{\text{AB}}=4\hat{\text{i}}+6\hat{\text{j}}+(\lambda+1)\hat{\text{k}}$
$\vec{\text{AC}}=3\hat{\text{i}}+10\hat{\text{j}}+5\hat{\text{k}}$
$\vec{\text{AD}}=4\hat{\text{i}}+5\hat{\text{j}}+5\hat{\text{k}}$
$\begin{vmatrix}4&6&(\lambda+1)\\3&10&5\\-4&5&5 \end{vmatrix}=0$
$4(50-25)-6(15+20)+(\lambda+1)(15+40)=0$
$100-210+55+55\lambda=0$
$55\lambda=55$
$\lambda=1$

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