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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
If the straight lines $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{\text{k}}=\frac{\text{z}}{2}$ and $\frac{\text{x}+1}{2}=\frac{\text{y}+1}{2}=\frac{\text{z}}{\text{k}}$ are coplanar, find the equation of the planes containing them.
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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 $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{\text{k}}=\frac{\text{z}}{2}$ and $\frac{\text{x}+1}{2}=\frac{\text{y}+1}{2}=\frac{\text{z}}{\text{k}}$ are coplanar.
$\therefore\ \begin{vmatrix} -1-1&-1-(-1)&0-0\\2&\text{k}&2\\2&2&\text{k}\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix} -2&0&0\\2&\text{k}&2\\2&2&\text{k}\end{vmatrix}=0$
$\Rightarrow-2(\text{k}^2-4)-0+0=0$
$\Rightarrow\text{k}^2-4=0$
$\Rightarrow\text{k}=\pm2$
The equation of the plane containing the given lines is $\begin{vmatrix} \text{x}-1&\text{y}+1&\text{z}\\2&\text{k}&2\\2&2&\text{k}\end{vmatrix}=0$
For k = 2, $\begin{vmatrix} \text{x}-1&\text{y}+1&\text{z}\\2&\text{k}&2\\2&2&\text{k}\end{vmatrix}=\begin{vmatrix} \text{x}-1&\text{y}+1&\text{z}\\2&\text{2}&2\\2&2&\text{2}\end{vmatrix}=0$
So, no plane exists for k = 2
For k = -2
$\begin{vmatrix} \text{x}-1&\text{y}+1&\text{z}\\2&\text{k}&2\\2&2&\text{k}\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix} \text{x}-1&\text{y}+1&\text{z}\\2&-\text{2}&2\\2&2&-\text{2}\end{vmatrix}=0$
$\Rightarrow(\text{x})(4-4)-(\text{y}+1)(-4-4)+\text{z}(4+4)=0$
$\Rightarrow8(\text{y}+1)+8\text{z}=0$
$\Rightarrow\text{y}+\text{z}+1=0$
Thus, the equation of the plane containing the given lines is y + z + 1 = 0

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