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# Find equation of plane containing two parallel lines $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{3}$ and $\dfrac{x}{4}=\dfrac{y-2}{-2}=\dfrac{z+1}{6}$ . Also find if the plane thus obtained contains the line $\dfrac{x-2}{3}=\dfrac{y-1}{1}=\dfrac{z-2}{5}$ or not.

Last updated date: 20th Jun 2024
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Hint: The equation of any line passing through a fixed point (a,b,c) and having the direction ratios as (l,m,n) is given as:
$\dfrac{x-a}{l}=\dfrac{y-b}{m}=\dfrac{z-c}{n}=\lambda$
Another important formula to solve this question is as follows:
The equation of a plane passing through a fixed point (a,b,c) is given as:
$l\left( x-a \right)+m\left( y-b \right)+n\left( z-c \right)=0$
(where l,m and n are the x-,y- and z-components of the equation normal to the plane)
Now, in this question, we will first find the normal vector for the unknown plane by finding the cross product of the direction ratios of the two parallel lines given with the direction ratios of another vector lying in the same plane which will be calculated by the fixed points through which our given two lines pass using the formula:
Direction ratios of a line passing through two fixed points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given by $\left( {{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}},{{z}_{2}}-{{z}_{1}} \right)$
This will give us the normal vector to the plane. Once we find the normal vector to the plane, we can use the above mentioned formula for finding the equation of the plane and hence we will get the equation of the required plane.

As mentioned in the question, we have to find the equation of the plane that contains both the given equations and along with this we have to find that whether the third line that is given lies in the plane or not.
Now, we have been given that the plane passes through two parallel lines whose equation are given as $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{3}$ and $\dfrac{x}{4}=\dfrac{y-2}{-2}=\dfrac{z+1}{6}$
Now, let the line $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{3}$ be ${{L}_{1}}$ and the line $\dfrac{x}{4}=\dfrac{y-2}{-2}=\dfrac{z+1}{6}$ be ${{L}_{2}}$.
The direction ratios of the line ${{L}_{1}}$ is (2,-1,3) and that of ${{L}_{2}}$ is (4,-2,6). The direction ratios of the line ${{L}_{2}}$ when further simplified will come as:
(4,-2,6)=(2,-1,3)
Now, this is the same as the direction ratios of ${{L}_{1}}$. We know that two lines with equal direction ratios are parallel to each other which just prove the already given condition in the question that these two lines are parallel.
Now we know that these two lines lie in the required plane, hence the vector parallel to the direction ratios of these lines will also lie in this plane.
The direction ratios of the lines lying in the plane are (2,-1,3).
Thus the vector parallel to them lying in the plane is given by:
$\vec{r}=2\hat{i}-\hat{j}+3\hat{k}$
Now again, the equations of the lines ${{L}_{1}}$ and ${{L}_{2}}$ lying in the plane are given by$\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{3}$ and $\dfrac{x}{4}=\dfrac{y-2}{-2}=\dfrac{z+1}{6}$ respectively.
From the equations of the lines, it is clear that the fixed point through which the line ${{L}_{1}}$ passes through is (1,-1,0) and the fixed point through which the line ${{L}_{2}}$ passes through is (0,2,-1).
Since these two lines lie in the requires plane, these two points will lie in the same plane as well.
Now, we know that direction ratios of a line passing through two fixed points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given as:
$\left( {{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}},{{z}_{2}}-{{z}_{1}} \right)$
Thus, the direction ratios of the line passing through the points (1,-1,0) and (0,2,-1) are given by:
\begin{align} & \left( 0-1,2-\left( -1 \right),-1-0 \right) \\ & \Rightarrow \left( -1,3,-1 \right) \\ \end{align}
Dividing it by -1 we get them as:
$\Rightarrow \left( 1,-3,1 \right)$
Now, we have already established that these two points lie in the plane. Hence, the line passing through these points will also lie in this plane and the vector parallel to the direction ratios of that point will also lie in the plane.
Hence, the second vector lying in the plane is given by:
$\vec{r}=\hat{i}-3\hat{j}+\hat{k}$
Now, we know that if a line passes through two vectors, the cross-product of those two vectors will give us the normal vector.
Now, we know that the cross product of two vectors ${{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and ${{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ is given as:
$\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ \end{matrix} \right|$
Hence, the normal vector is given by:
\begin{align} & \vec{n}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & -1 & 3 \\ 1 & -3 & 1 \\ \end{matrix} \right| \\ & \vec{n}=\hat{i}\left( -1+9 \right)-\hat{j}\left( 2-3 \right)+\hat{k}\left( -6+1 \right) \\ & \vec{n}=8\hat{i}+\hat{j}-5\hat{k} \\ \end{align}
Now, we know that the equation of a plane passing through a fixed point (a,b,c) is given as:
$l\left( x-a \right)+m\left( y-b \right)+n\left( z-c \right)=0$
(where l,m and n are the x-,y- and z-components of the equation normal to the plane)
We can here take any one the fixed points obtained earlier. Let us take that to be (1,-1,0)
Thus, the equation of our plane will come out as:
$8\left( x-1 \right)+1\left( y-\left( -1 \right) \right)+\left( -5 \right)\left( z-0 \right)=0$
Simplifying it, we will get:
\begin{align} & 8x-8+y+1-5z=0 \\ & \Rightarrow 8x+y-5z=7 \\ \end{align}
Hence, the required equation of the plane is $8x+y-5z=7$

Note: We can also find the equation of the line by using scalar triple product. Once, we find two vectors lying in the plane, we can assume a variable point in the plane as (x,y,z) and then we can take any of the fixed points and find the direction ratio of the line passing through these points. Let us take that point as (1,-1,0).
Thus, the direction ratios will be:
$(x-1,y+1,z)$
Thus, the vector parallel to it will be:
$\vec{r}=\left( x-1 \right)\hat{i}+\left( y+1 \right)\hat{j}+z\hat{k}$
So, the three vectors $\left( x-1 \right)\hat{i}+\left( y+1 \right)\hat{j}+z\hat{k}$, $\hat{i}-3\hat{j}+\hat{k}$and $2\hat{i}-\hat{j}+3\hat{k}$ will all lie in the required plane and hence they will be coplanar.
Now, we know that the STP of three coplanar vectors is 0.
Thus, using doing the STP of these three vectors we get:
\begin{align} & \left| \begin{matrix} x-1 & y+1 & z \\ 1 & -3 & 1 \\ 2 & -1 & 3 \\ \end{matrix} \right|=0 \\ & \Rightarrow \left( x-1 \right)\left( -9+1 \right)-\left( y+1 \right)\left( 3-2 \right)+z\left( -1+6 \right)=0 \\ & \Rightarrow -8x+8-y-1+5z=0 \\ & \Rightarrow 8x+y-5z=7 \\ \end{align}