Question

If ${{L}_{1}}$ is the line of intersection of the planes $2x-2y+3z-2=0$, $x-y+z+1=0$ and ${{L}_{2}}$ is the line of intersection of the planes $x+2y-z-3=0$, $3x-y+2z-1=0$, then the distance of origin from the plane containing lines ${{L}_{1}}$ and ${{L}_{2}}$ is

Answer 1

We solve this question by first finding the direction ratios of line ${{L}_{1}}$ from the direction ratios of given planes. Then we find the direction ratios of line ${{L}_{2}}$ using the other two given planes direction ratios. Then we find the direction ratios of line normal to the plane containing both the lines by finding the cross product of direction ratios of lines ${{L}_{1}}$ and ${{L}_{2}}$. Then we use the formula for the equation of the plane having normal direction ratios using formula $ax+by+cz+d=0$. Then we use any point on line ${{L}_{1}}$ and substitute it in the plane equation obtained to find the value of d. Then we substitute the obtained d value to get the required equation of plane.

Complete step-by-step answer:
Let us assume that the direction ratios of ${{L}_{1}}$ are $\left( l,m,n \right)$ and ${{L}_{2}}$ are $\left( p,q,r \right)$.
As ${{L}_{1}}$ is the intersection of the planes $2x-2y+3z-2=0$ and $x-y+z+1=0$, it is present in both the planes. Then the angle between the line and the plane is zero.
Now, let us consider the formula for angle between line with direction ratios $\left( l,m,n \right)$ and plane $ax+by+cz+d=0$
$\sin \theta =\dfrac{al+bm+cn}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}$
When the angle between them is zero,
$\begin{aligned} & \sin {{0}^{\circ }}=0=\dfrac{al+bm+cn}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}} \\\ & al+bm+cn=0 \\\ \end{aligned}$
So, applying above formula we get that
$\begin{aligned} & 2l-2m+3n=0 \\\ & l-m+n=0 \\\ \end{aligned}$
Solving them we get,
$\begin{aligned} & \Rightarrow \dfrac{l}{\left| \begin{matrix} -2 & 3 \\\ -1 & 1 \\\ \end{matrix} \right|}=\dfrac{m}{-\left| \begin{matrix} 2 & 3 \\\ 1 & 1 \\\ \end{matrix} \right|}=\dfrac{n}{\left| \begin{matrix} 2 & -2 \\\ 1 & -1 \\\ \end{matrix} \right|} \\\ & \Rightarrow \dfrac{l}{\left( -2+3 \right)}=\dfrac{m}{-\left( 2-3 \right)}=\dfrac{n}{\left( -2+2 \right)} \\\ & \Rightarrow \dfrac{l}{1}=\dfrac{m}{1}=\dfrac{n}{0} \\\ \end{aligned}$
So, we get the direction ratios of ${{L}_{1}}$ as $\left( 1,1,0 \right)$.
Now let us consider the second line.
As ${{L}_{2}}$ is the intersection of the planes $x+2y-z-3=0$ and $3x-y+2z-1=0$, it is present in both the planes. Then the angle between the line and the plane is zero.
Using the above discussed formula when angle between line and plane is zero,
$al+bm+cn=0$
So, applying above formula we get that
$\begin{aligned} & p+2q-r=0 \\\ & 3p-q+2r=0 \\\ \end{aligned}$
Solving them we get,
$\begin{aligned} & \Rightarrow \dfrac{p}{\left| \begin{matrix} 2 & -1 \\\ -1 & 2 \\\ \end{matrix} \right|}=\dfrac{q}{-\left| \begin{matrix} 1 & -1 \\\ 3 & 2 \\\ \end{matrix} \right|}=\dfrac{r}{\left| \begin{matrix} 1 & 2 \\\ 3 & -1 \\\ \end{matrix} \right|} \\\ & \Rightarrow \dfrac{p}{\left( 4-1 \right)}=\dfrac{q}{-\left( 2+3 \right)}=\dfrac{r}{\left( -1-6 \right)} \\\ & \Rightarrow \dfrac{p}{3}=\dfrac{q}{-5}=\dfrac{r}{-7} \\\ \end{aligned}$
So, we get the direction ratios of ${{L}_{2}}$ as $\left( 3,-5,-7 \right)$.
We need to find the plane containing the lines ${{L}_{1}}$ and ${{L}_{2}}$. So, any line normal to the plane is normal to the lines ${{L}_{1}}$ and ${{L}_{2}}$.
Now let us consider the concept that any line perpendicular to two lines a and b is of the form $a\times b$.
So, using this formula, normal to the required plane have direction ratios ${{L}_{1}}\times {{L}_{2}}$.
So, by finding the cross product of direction ratios of ${{L}_{1}}$ and ${{L}_{2}}$, we get,
$\begin{aligned} & \Rightarrow \left| \begin{matrix} i & j & k \\\ 1 & 1 & 0 \\\ 3 & -5 & -7 \\\ \end{matrix} \right|=i\left| \begin{matrix} 1 & 0 \\\ -5 & -7 \\\ \end{matrix} \right|-j\left| \begin{matrix} 1 & 0 \\\ 3 & -7 \\\ \end{matrix} \right|+k\left| \begin{matrix} 1 & 1 \\\ 3 & -5 \\\ \end{matrix} \right| \\\ & \Rightarrow \left| \begin{matrix} i & j & k \\\ 1 & 1 & 0 \\\ 3 & -5 & -7 \\\ \end{matrix} \right|=-7i+7j-8k \\\ \end{aligned}$
So, we get the direction ratios of normal as $\left( -7,7,-8 \right)$.
Let us use the formula for equation of plane with normal having direction ratios $\left( a,b,c \right)$ is
$ax+by+cz+d=0$
Using the above formula equation of plane with normal having direction ratios $\left( -7,7,-8 \right)$ is
$\Rightarrow -7x+7y-8z+d=0$
As line ${{L}_{1}}$ is in the plane any point on the line is on the plane. So, let us find any point on the line ${{L}_{1}}$.
As ${{L}_{1}}$ is the intersection of the planes $2x-2y+3z-2=0$ and $x-y+z+1=0$, any point on this line satisfies both the plane equations.
Let us assume the point with x=0. Then
$\begin{aligned} & \Rightarrow 2x-2y+3z-2=0 \\\ & \Rightarrow -2y+3z-2=0 \\\ \end{aligned}$
$\begin{aligned} & \Rightarrow x-y+z+1=0 \\\ & \Rightarrow -y+z+1=0 \\\ & \Rightarrow y=z+1 \\\ \end{aligned}$
Substituting this value of y in above equation we get,
$\begin{aligned} & \Rightarrow -2\left( z+1 \right)+3z-2=0 \\\ & \Rightarrow -2z-2+3z-2=0 \\\ & \Rightarrow z-4=0 \\\ & \Rightarrow z=4 \\\ \end{aligned}$
So, using this value we can get value of y as,
$\Rightarrow y=z+1=4+1=5$
So, we get that the point $\left( 0,5,4 \right)$ is on the line ${{L}_{1}}$. So, it must satisfy the equation of the plane containing ${{L}_{1}}$ and ${{L}_{2}}$. So, substituting the point in the equation of plane we get,
$\begin{aligned} & \Rightarrow -7x+7y-8z+d=0 \\\ & \Rightarrow -7\left( 0 \right)+7\left( 5 \right)-8\left( 4 \right)+d=0 \\\ & \Rightarrow 0+35-32+d=0 \\\ & \Rightarrow d+3=0 \\\ & \Rightarrow d=-3 \\\ \end{aligned}$
So, substituting this value in the equation of the plane we get,
$\begin{aligned} & \Rightarrow -7x+7y-8z-3=0 \\\ & \Rightarrow 7x-7y+8z+3=0 \\\ \end{aligned}$
Hence the equation of the plane containing lines ${{L}_{1}}$ and ${{L}_{2}}$ is $7x-7y+8z+3=0$.
Hence, the answer is $7x-7y+8z+3=0$.

Note: The common mistake that one makes while solving this problem is one might get confused and take the formula for the angle between the line and plane as
$\cos \theta =\dfrac{al+bm+cn}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}$
But it is wrong. We take cosine in the formula for angle between two lines and sine for angle between line and plane.