Question

The shortest distance between the y-axis and the line $2x + 3y + 5z + 1 = 3x + 4y + 6z + 2 = 0$ is $\dfrac{2}{{\sqrt k }},$ then the value of k is

Answer 1

Hint: Form the equation of the line that contains both the planes (intersection of planes) the perpendicular distance between two lines i.e; $y = 0$ and the required line gives us the shortest distance

Complete step by step solution:
Let ${{\text{P}}_1}:\;\;\;\;\;2x + 3y + 5z + 1 = 0$
     ⇒ (2, 3, 5) are direction ratios of ${{\text{P}}_1}$
Similarly (3, 4, 6) are direction ratios of ${{\text{P}}_2}$
Where \[{{\text{P}}_2}:\;\;\;\;\;\;\;3x + 4y + 6z + 2 = 0\]
Since the line of intersection of ${{\text{P}}_1}\;\& \;{{\text{P}}_2}$ is perpendicular to the normal of ${{\text{P}}_1}\;\& \;{{\text{P}}_2}$
The Direction ratios of the line can be found out with the real cross product of d r s of ${{\text{P}}_1}\;\& {{\text{P}}_2}$
d. r’s of $L = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  2&3&5 \\
  3&4&6
\end{array}} \right|$
if the plane line on y-axis then equation of line is
$y = 0$ in ${{\text{P}}_1}$: $2x + 5z = - 1$ (1)
$y = 0$ in ${{\text{P}}_2}$ $3x + 6z = - 2$ (2)
On solving eqns (1) & (2)
We get $3z = 1$ or$z = \dfrac{1}{3}$
and $2x = \dfrac{8}{3}$ or$x = - \dfrac{4}{3}$
we can now from the eqn of line as we know the d.r’s and a pt that line
$p: - \left( {\dfrac{{ - 4}}{3},0,\dfrac{1}{3}} \right)$ d.r’n (−2, −3, −1)
equation of line is
$\dfrac{{x - {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z - {z_1}}}{n}$
$e{q^n}$ of L:-$\dfrac{{x - \left( {\dfrac{{ - 4}}{3}} \right)}}{2} = \dfrac{{y - 0}}{{ - 3}} = \dfrac{{z\dfrac{{ - 1}}{3}}}{{ - 1}}$
Any point on y axis is given by:-(0,y,0)
Then other line $e{q^n}$ is :-$\dfrac{x}{0} = \dfrac{y}{1} = \dfrac{z}{0}$
The shortest distance (d) formula is given by
$ = \dfrac{{\left\| {\begin{array}{*{20}{c}}
  {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}}
\end{array}} \right\|}}{{\left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}}
\end{array}} \right|}}$
Where $\left( {{a_i},\;{b_i},\;{c_i}} \right)$ are direction vatias and $\left( {{x_i},\;{y_i}\;{z_i}} \right)$ are points on respective lines.
$Therefore, \;d = \dfrac{{\left\| {\begin{array}{*{20}{c}}
  {\dfrac{4}{3}}&1&{\dfrac{{ - 1}}{3}} \\
  { - 2}&{ - 3}&{ - 1} \\
  0&1&0
\end{array}} \right\|}}{{\left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  { - 2}&{ - 3}&{ - 1} \\
  0&1&0
\end{array}} \right|}}$
\[d = \dfrac{{\left| {\dfrac{4}{3}\left( 1 \right) - 1\left( 0 \right) - \dfrac{1}{3}\left( { - 2} \right)} \right|}}{{\left| {\begin{array}{*{20}{c}}
  {o\hat i}&{ - o\hat j}&{ - 2\hat k}
\end{array}} \right|}}\]
$d = \left( {\dfrac{6}{3}} \right) \times \dfrac{1}{{\sqrt 5 }}$
$d = \dfrac{2}{{\sqrt 5 }}$

Hence value of k is $\sqrt k $

Note: Intersection of two planes is always a line, not a point. The shortest distance between two lines is always the perpendicular distance. This question can also be solved using a family of lines that pass through a given plane.