Question

The perpendicular distance from the origin to the plane containing the two lines, \[\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}\] and \[\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}\].
\[\begin{align}
  & A)\dfrac{11}{\sqrt{6}} \\
 & B)6\sqrt{11} \\
 & C)\sqrt{11} \\
 & D)11\sqrt{6} \\
\end{align}\]

Answer 1

Hint: We know that the plane containing the two lines, \[\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}\] and \[\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}\] is \[\left| \begin{matrix}
   x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
\end{matrix} \right|=0\]. We know that the distance between origin and \[ax+by+cz+d=0\] is equal to \[\dfrac{\left| d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\]. By using these concepts, we can get the perpendicular distance from the origin to the plane containing the two lines, \[\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}\] and \[\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}\].

Complete step-by-step solution:
Now we should find the plane containing the two lines, \[\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}\] and \[\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}\].
We know that the plane containing the two lines, \[\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}\] and \[\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}\] is \[\left| \begin{matrix}
   x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
\end{matrix} \right|=0\].
By using the above concept, we can find the plane containing the two lines, \[\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}\] and \[\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}\].
Let us assume this plane containing the two lines, \[\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}\] and \[\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}\] as plane P.
\[\Rightarrow P\equiv \left| \begin{matrix}
   x+2 & y-2 & z+5 \\
   3 & 5 & 7 \\
   1 & 4 & 7 \\
\end{matrix} \right|=0\]
We know that the value of \[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=a\left( ei-hf \right)-b\left( di-fg \right)+c\left( dh-ge \right)\].
Now by using this concept, we should find the equation of plane P.
\[\begin{align}
  & \Rightarrow P\equiv \left| \begin{matrix}
   x+2 & y-2 & z+5 \\
   3 & 5 & 7 \\
   1 & 4 & 7 \\
\end{matrix} \right|=0 \\
 & \Rightarrow P\equiv \left( x+2 \right)\left( (5)(7)-(4)(7) \right)-\left( y-2 \right)\left( (3)(7)-(7)(1) \right)+\left( z+5 \right)\left( (3)(7)-(1)(7) \right)=0 \\
 & \Rightarrow P\equiv \left( x+2 \right)\left( 35-28 \right)-\left( y-2 \right)\left( 21-7 \right)+\left( z+5 \right)\left( 21-14 \right)=0 \\
 & \Rightarrow P\equiv \left( x+2 \right)\left( 7 \right)-\left( y-2 \right)\left( 14 \right)+\left( z+5 \right)\left( 7 \right)=0 \\
 & \Rightarrow P\equiv 7x+14-14y+28+7z+35=0 \\
 & \Rightarrow P\equiv 7x-14y+7z+77=0 \\
 & \Rightarrow P\equiv 7\left( x-2y+z+11 \right)=0 \\
 & \Rightarrow P\equiv x-2y+z+11=0.....(1) \\
\end{align}\]
From equation (1), it is clear that \[x-2y+z+11=0\] is the equation of plane containing the two lines, \[\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}\] and \[\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}\].
Now we should find the distance between origin and \[x-2y+z+11=0\].
We know that the distance between origin and \[ax+by+cz+d=0\] is equal to \[\dfrac{\left| d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\].
Now by using this concept, we can find the distance between origin and \[x-2y+z+11=0\].
Let us assume this distance is equal to D.
\[\begin{align}
  & \Rightarrow D=\dfrac{\left| 11 \right|}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{1}^{2}}}} \\
 & \Rightarrow D=\dfrac{11}{\sqrt{6}}....(2) \\
\end{align}\]
From equation (2), it is clear that the distance between origin and \[x-2y+z+11=0\] is equal to \[\dfrac{11}{\sqrt{6}}\].
Hence, option A is correct.

Note: Students may have a misconception that the distance between origin and \[ax+by+cz+d=0\] is equal to \[\dfrac{\left| d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}}}\]. If this misconception is followed, then we cannot get the exact value of D. So, this misconception should be avoided. Otherwise, we cannot get correct answer at any cost.