Question

If ${L_1}$ is the line of intersection of the plane $2x - 2y + 3z - 2 = 0$, $x - y + z + 1 = 0$ and ${L_2}$ is the line of intersection of the planes $x + 2y - z - 3 = 0$, $3x - y + 2z - 1 = 0$. Find the distance of the origin from the plane, containing the lines ${L_1}$ and ${L_2}$.
(A) $\dfrac{1}{{2\sqrt 2 }}$
(B) $\dfrac{1}{{\sqrt 2 }}$
(C) $\dfrac{1}{{4\sqrt 2 }}$
(D) $\dfrac{1}{{3\sqrt 2 }}$

Answer 1

Hint: Since given lines lie in two planes, so the direction vector of the line will be perpendicular to normal vectors of the plane. Use this to find vectors for lines. Now using any two planes, find a point lying on the line. Take the cross product of direction vectors of lines to find a normal vector to the required plane. Now you’ll have a normal vector and a point on the plane. Using these two, find the vector equation of the plane. Now calculate the distance from the point $\left( {0,0,0} \right)$to the plane.

Complete step-by-step answer:
According to the given information the plane $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$ intersect at a line ${L_1}$. And planes $x + 2y - z - 3 = 0$ and $3x - y + 2z - 1 = 0$ intersect at the line ${L_2}$.
Since the line of the intersection will lie in both planes, it will be perpendicular to both normal vectors. We can take the cross product of the normal vectors to find the direction vector of the line of intersection of the planes.
$ \Rightarrow $ The direction vector for line ${L_1}$, i.e. the line ${L_1}$ is parallel to: $\left| {\begin{array}{*{20}{c}}
  {\widehat i}&{\widehat j}&{\widehat k} \\
  2&{ - 2}&3 \\
  1&{ - 1}&1
\end{array}} \right| = \widehat i\left( { - 2 + 3} \right) + \widehat j\left( { - 2 + 3} \right) + \widehat k\left( 0 \right) = \widehat i + \widehat j$
Similarly, the direction vector of line ${L_2}$, i.e. the line ${L_2}$ is parallel to: $\left| {\begin{array}{*{20}{c}}
  {\widehat i}&{\widehat j}&{\widehat k} \\
  1&2&{ - 1} \\
  3&{ - 1}&2
\end{array}} \right| = \widehat i\left( {4 - 1} \right) + \widehat j\left( { - 2 - 3} \right) + \widehat k\left( { - 1 - 6} \right) = 3\widehat i - 5\widehat j - 7\widehat k$
Let us now find a point that lies in the same plane as the two lines ${L_1}$ and ${L_2}$. We can do that by finding any point that lies on the line ${L_1}$ or ${L_2}$, i.e. on the line of intersection of planes.
Using the equations of the plane $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$, we can put $x = 0$ to get equations as:
$ \Rightarrow - 2y + 3z - 2 = 0{\text{ and }} - y + z + 1 = 0$ (1) and (2)
Now, we can substitute the value of $y$ from the equation (2) into the equation (1).
So, (1) will become: $ - 2\left( {z + 1} \right) + 3z - 2 = 0 \Rightarrow 3z - 2z - 2 - 2 = 0 \Rightarrow z = 4$
Putting $z = 4$ back in the equation (2), we get: $y = z + 1 = 4 + 1 = 5$
Therefore, we calculated a point that lies on the plane that carries the line ${L_1}$ and ${L_2}$ as $\left( {0,5,4} \right)$.
The cross product of two vectors will give a vector that is normal to the plane containing those two vectors. But normalizing a vector means to convert it into a unit vector. In your case, a normalized vector of the cross product will give you a unit vector.
$ \Rightarrow $ For the normal vector $\vec n$ of the plane containing lines ${L_1}$ and ${L_2}$ can be written as: $\vec n = {\vec L_1} \times {\vec L_2}$
\[ \Rightarrow \vec n = {\vec L_1} \times {\vec L_2} = \left| {\begin{array}{*{20}{c}}
  {\widehat i}&{\widehat j}&{\widehat k} \\
  1&1&0 \\
  3&{ - 5}&{ - 7}
\end{array}} \right| = \widehat i\left( { - 7 - 0} \right) + \widehat j\left( {7 + 0} \right) + \widehat k\left( { - 5 - 3} \right) = - 7\widehat i + 7\widehat j - 8\widehat k\]
Consider the plane passing through the point ${P_0}\left( {{x_0},{y_0},{z_0}} \right)$ with the normal vector\[\overrightarrow n \]. Let $\overrightarrow {{r_0}} $ be the position vector of point ${P_0}\left( {{x_0},{y_0},{z_0}} \right)$. Then the vector equation of the plane is
\[ \Rightarrow \overrightarrow n \cdot \left( {\overrightarrow r - \overrightarrow {{r_0}} } \right) = 0{\text{ or }}\overrightarrow n \cdot \overrightarrow r = \overrightarrow n \cdot {\overrightarrow r _0}\]
We already have point ${P_0}\left( {{x_0},{y_0},{z_0}} \right) = \left( {0,5,4} \right)$ and normal vector as \[\vec n = - 7\widehat i + 7\widehat j - 8\widehat k\]
Equation of the plane can be written as:
\[ \Rightarrow \left( { - 7\widehat i + 7\widehat j - 8\widehat k} \right) \cdot \left( {x\widehat i + y\widehat j + z\widehat k} \right) = \left( { - 7\widehat i + 7\widehat j - 8\widehat k} \right) \cdot \left( {5\widehat j + 4\widehat k} \right)\]
Now, we can solve this dot product to find the equation of a plane in Cartesian form
 \[ \Rightarrow \left( { - 7 \times x} \right) + \left( {7 \times y} \right) + \left( { - 8 \times z} \right) = \left( { - 7 \times 0} \right) + \left( {7 \times 5} \right) + \left( { - 8 \times 4} \right)\]
\[ \Rightarrow - 7x + 7y - 8z = 35 - 32 = 3\]
Therefore, we got the equation of the required plane as \[7x - 7y + 8z + 3 = 0\]
Now, we need to find the distance of the origin from this plane.
As we know that the distance of the point $\left( {{x_1},{y_1},{z_1}} \right)$ from the plane $Ax + By + Cz + D = 0$ is $\left| {\dfrac{{A{x_1} + B{y_1} + C{z_1} + D}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right|$.
 But for origin, the point $\left( {{x_1},{y_1},{z_1}} \right) = \left( {0,0,0} \right)$ and thus the distance will be $\left| {\dfrac{D}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right|$
$ \Rightarrow $ Distance from origin to the plane $ = \left| {\dfrac{3}{{\sqrt {{7^2} + {{\left( { - 7} \right)}^2} + {8^2}} }}} \right| = \left| {\dfrac{3}{{\sqrt {49 + 49 + 64} }}} \right| = \dfrac{3}{{\sqrt {162} }} = \dfrac{1}{{\sqrt {18} }} = \dfrac{1}{{3\sqrt 2 }}$

So, the correct answer is “Option D”.

Note: Follow a step by step procedure while solving the problem. Notice that we used cross product $\left( \times \right)$ for finding normal vectors while dot or scalar product $\left( \cdot \right)$ for finding the equation of the plane. The cross product $\vec a \times \vec b$ is defined as a vector $\vec c$ that is perpendicular (orthogonal) to both $\vec a{\text{ and }}\vec b$, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.