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Scalar Triple Product — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsScalar Triple Product3 Marks
Question
Find the value of $\lambda$ so that the following vectors are coplanar:
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\vec{\text{c}}=\lambda\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}$
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Answer

Given:
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{c}}=\lambda\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}$
We know that vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar if $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0$
It is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar.
$\therefore\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0$
$\Rightarrow\begin{vmatrix}1&-1&1\\2&1&-1\\\lambda&-1&\lambda \end{vmatrix}=0$
$\Rightarrow 1(\lambda-1)+1(2\lambda+\lambda)+1(-2-\lambda)=0$
$\Rightarrow \lambda -1+3\lambda-2-\lambda=0$
$\Rightarrow 3\lambda-3=0$
$\Rightarrow\lambda=1$

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