Question
Show that lines $\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})$ and $\bar{r}=(4 \hat{i}-3 \hat{j}+2 \hat{k})+\mu(\hat{i}-2 \hat{j}+2 \hat{k})$ are coplanar. Find the equation of the plane determined by them.
✓
Answer
Lines $\bar{r}=\bar{a}_1+\lambda_1 \bar{b}_1$ and $\bar{r}=\bar{a}_2+\lambda_2 \bar{b}_2$ are coplanar if and only if $\left(\bar{a}_2-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=0$
Here $\bar{a}_1=\hat{i}+\hat{j}-\hat{k}, \bar{a}_2=4 \hat{i}-3 \hat{j}+2 \hat{k}$
$
\begin{aligned}
& \bar{b}_1=2 \hat{i}-2 \hat{j}+\hat{k}, \bar{b}_2=\hat{i}-2 \hat{j}+2 \hat{k} \\
& \therefore \bar{a}_2-\bar{a}_1=3 \hat{i}-4 \hat{j}+3 \hat{k} \\
& \left(\bar{a}_2-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=\left|\begin{array}{lll}
3 & -4 & 3 \\
2 & -2 & 1 \\
1 & -2 & 2
\end{array}\right|=3(-2)+4(3)+3(-2)=-6+12-6=0
\end{aligned}
$
$\therefore$ Given lines are coplanar.
Now $\bar{b}_1 \times \bar{b}_2=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -2 & 2\end{array}\right|=-2 \hat{i}-3 \hat{j}-2 \hat{k}$
The equation of the plane determined by them is $\left(\bar{r}-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=0$
$
\begin{aligned}
& \therefore \bar{r} \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=\bar{a}_1 \cdot\left(\bar{b}_1 \times \bar{b}_2\right) \\
& \therefore \bar{r} \cdot(-2 \hat{i}-3 \hat{j}-2 \hat{k})=(\hat{i}+\hat{j}-\hat{k}) \cdot(-2 \hat{i}-3 \hat{j}-2 \hat{k}) \\
& \therefore \bar{r} \cdot(-2 \hat{i}-3 \hat{j}-2 \hat{k})=-3 \\
& \therefore \bar{r} \cdot(2 \hat{i}+3 \hat{j}+2 \hat{k})=3
\end{aligned}
$
Here $\bar{a}_1=\hat{i}+\hat{j}-\hat{k}, \bar{a}_2=4 \hat{i}-3 \hat{j}+2 \hat{k}$
$
\begin{aligned}
& \bar{b}_1=2 \hat{i}-2 \hat{j}+\hat{k}, \bar{b}_2=\hat{i}-2 \hat{j}+2 \hat{k} \\
& \therefore \bar{a}_2-\bar{a}_1=3 \hat{i}-4 \hat{j}+3 \hat{k} \\
& \left(\bar{a}_2-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=\left|\begin{array}{lll}
3 & -4 & 3 \\
2 & -2 & 1 \\
1 & -2 & 2
\end{array}\right|=3(-2)+4(3)+3(-2)=-6+12-6=0
\end{aligned}
$
$\therefore$ Given lines are coplanar.
Now $\bar{b}_1 \times \bar{b}_2=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -2 & 2\end{array}\right|=-2 \hat{i}-3 \hat{j}-2 \hat{k}$
The equation of the plane determined by them is $\left(\bar{r}-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=0$
$
\begin{aligned}
& \therefore \bar{r} \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=\bar{a}_1 \cdot\left(\bar{b}_1 \times \bar{b}_2\right) \\
& \therefore \bar{r} \cdot(-2 \hat{i}-3 \hat{j}-2 \hat{k})=(\hat{i}+\hat{j}-\hat{k}) \cdot(-2 \hat{i}-3 \hat{j}-2 \hat{k}) \\
& \therefore \bar{r} \cdot(-2 \hat{i}-3 \hat{j}-2 \hat{k})=-3 \\
& \therefore \bar{r} \cdot(2 \hat{i}+3 \hat{j}+2 \hat{k})=3
\end{aligned}
$
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