Question

If the distance between the plane \[Ax - 2y + z = d\] and the plane containing the lines \[\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{4}\] and \[\dfrac{{x - 2}}{3} = \dfrac{{y - 3}}{4} = \dfrac{{z - 4}}{5}\] is \[\sqrt 6 \], then the value of \[\left| d \right|\] is
A. 3
B. 4
C. 5
D. 6

Answer 1

Hint: First of all, find the equation of the plane in which the given two lines are containing. Then use the formula that the distance between the two planes \[{a_1}x + {b_1}y + {c_1}z = {d_1}\] and \[{a_2}x + {b_2}y + {c_2}z = {d_2}\] is given by \[\dfrac{{\left| {{d_2} - {d_1}} \right|}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} }}\] to get the required answer.

Complete step-by-step answer:
Let the given plane equation is \[{P_1}:Ax - 2y + z = d\]
Let \[{P_2}\] be the equation of the plane containing the lines \[\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{4}\] and \[\dfrac{{x - 2}}{3} = \dfrac{{y - 3}}{4} = \dfrac{{z - 4}}{5}\].
We know that the plane equation containing the lines \[\dfrac{{x - {x_1}}}{{{p_1}}} = \dfrac{{y - {y_1}}}{{{p_2}}} = \dfrac{{z - {z_1}}}{{{p_3}}}\] and \[\dfrac{{x - {x_2}}}{{{q_1}}} = \dfrac{{y - {y_2}}}{{{q_2}}} = \dfrac{{z - {z_2}}}{{{q_3}}}\] is given by \[\left| {\begin{array}{*{20}{c}}
  {x - {x_1}}&{y - {y_1}}&{z - {z_1}} \\
  {{p_1}}&{{p_2}}&{{p_3}} \\
  {{q_1}}&{{q_2}}&{{q_3}}
\end{array}} \right| = 0\] or \[\left| {\begin{array}{*{20}{c}}
  {x - {x_2}}&{y - {y_2}}&{z - {z_2}} \\
  {{p_1}}&{{p_2}}&{{p_3}} \\
  {{q_1}}&{{q_2}}&{{q_3}}
\end{array}} \right| = 0\].
So, the equation of plane \[{P_2}\] is given by
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {x - 1}&{y - 2}&{z - 3} \\
  2&3&4 \\
  3&4&5
\end{array}} \right| = 0\]
Opening the determinant, we have
\[
   \Rightarrow \left( {x - 1} \right)\left[ {3 \times 5 - 4 \times 4} \right] - \left( {y - 2} \right)\left[ {2 \times 5 - 3 \times 4} \right] + \left( {z - 3} \right)\left[ {2 \times 4 - 3 \times 3} \right] = \\
   \Rightarrow \left( {x - 1} \right)\left( {15 - 16} \right) - \left( {y - 2} \right)\left( {10 - 12} \right) + \left( {z - 3} \right)\left( {8 - 9} \right) = 0 \\
   \Rightarrow - \left( {x - 1} \right) + 2\left( {y - 2} \right) - \left( {z - 3} \right) = 0 \\
   \Rightarrow - x + 1 + 2y - 4 - z + 3 = 0 \\
  \therefore {P_2}:x - 2y + z = 0 \\
\]
We know that the distance between the two planes \[{a_1}x + {b_1}y + {c_1}z = {d_1}\] and \[{a_2}x + {b_2}y + {c_2}z = {d_2}\] is given by \[\dfrac{{\left| {{d_2} - {d_1}} \right|}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} }}\].
Given that the distance between the two planes \[{P_1}\& {P_2}\] is \[\sqrt 6 \]. So, we have
\[
   \Rightarrow \sqrt 6 = \dfrac{{\left| {d - 0} \right|}}{{\sqrt {{1^2} + {{\left( { - 2} \right)}^2} + {1^2}} }} \\
   \Rightarrow \sqrt 6 = \dfrac{{\left| d \right|}}{{\sqrt {1 + 4 + 1} }} \\
   \Rightarrow \sqrt 6 \times \sqrt 6 = \left| d \right| \\
  \therefore \left| d \right| = 6 \\
\]
Thus, the correct option is D. 6

Note: The plane equation containing the lines \[\dfrac{{x - {x_1}}}{{{p_1}}} = \dfrac{{y - {y_1}}}{{{p_2}}} = \dfrac{{z - {z_1}}}{{{p_3}}}\] and \[\dfrac{{x - {x_2}}}{{{q_1}}} = \dfrac{{y - {y_2}}}{{{q_2}}} = \dfrac{{z - {z_2}}}{{{q_3}}}\] is given by \[\left| {\begin{array}{*{20}{c}}
  {x - {x_1}}&{y - {y_1}}&{z - {z_1}} \\
  {{p_1}}&{{p_2}}&{{p_3}} \\
  {{q_1}}&{{q_2}}&{{q_3}}
\end{array}} \right| = 0\] or \[\left| {\begin{array}{*{20}{c}}
  {x - {x_2}}&{y - {y_2}}&{z - {z_2}} \\
  {{p_1}}&{{p_2}}&{{p_3}} \\
  {{q_1}}&{{q_2}}&{{q_3}}
\end{array}} \right| = 0\].