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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry4 Marks
Question
Find the value of $\lambda$ for which the lines $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}+3}{\lambda^2}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}-2}{\lambda^2}=\frac{\text{z}-1}{2}$ are coplanar.
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Answer

The lines $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$ are coplanar if
$\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}=0$
The given lines $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}+3}{\lambda^2}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}-2}{\lambda^2}=\frac{\text{z}-1}{2}$ are coplanar.
$\therefore\ \begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}3-1&2-2&1-(-3)\\1&2&\lambda^2\\1&\lambda^2&2\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}2&0&4\\1&2&\lambda^2\\1&\lambda^2&2\end{vmatrix}=0$
$\Rightarrow2(4-\lambda^4)-0+4(\lambda^4-2)=0$
$\Rightarrow-2\lambda^4+4\lambda^2=0$
$\Rightarrow\lambda^2(\lambda^2-2)=0$
$\Rightarrow\lambda^2=0\text{ or }\lambda^2-2=0$
$\Rightarrow\lambda=0\text{ or }\lambda=\pm\sqrt{2}$
Thus, the values of $\lambda$ are $0,-\sqrt{2}$ and $\sqrt{2}$

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