Question

A line having direction ratios 3,4 ,5 cuts 2 planes $ 2x - 3y + 6z - 12 = 0 $ and $ 2x - 3y + 6z + 2 = 0 $ at point P & Q, then finds the length of PQ.

Answer 1

Hint: A plane is a flat surface having 2 dimensions; it extends to infinity in two opposite directions and doesn’t have any thickness. A plane can also be said to be generated by a straight line moving at a constant velocity with respect to a fixed point. Any point lying on a plane or a line has to satisfy its equation. Using this condition, we can find out the coordinates of the points P and Q, and thus the distance between them.

Complete step-by-step answer:
The line cutting the two planes has direction ratios 3, 4, 5 so any point on this line will have coordinates of the form $ (3a,4a,5a) $
Equation of plane one is $ 2x - 3y + 6z - 12 = 0 $
Any point lying on this plane will satisfy this equation. That is,
 $
  2(3a) - 3(4a) + 6(5a) - 12 = 0 \\
   \Rightarrow 6a - 12a + 30a - 12 = 0 \\
   \Rightarrow 24a = 12 \\
   \Rightarrow a = \dfrac{1}{2} \;
  $
So the coordinates of the point where the line cuts the plane is $ P\,(\dfrac{3}{2},2,\dfrac{5}{2}) $
Equation of plane two is $ 2x - 3y + 6z + 2 = 0 $
Any point lying on this plane will satisfy this equation. That is,
 $
  2(3a) - 3(4a) + 6(5a) + 2 = 0 \\
   \Rightarrow 6a - 12a + 30a + 2 = 0 \\
   \Rightarrow 24a = - 2 \\
   \Rightarrow a = \dfrac{{ - 1}}{{12}} \;
  $
So the coordinates of the point where the line cuts the plane is
$ Q\,(\dfrac{{ - 1}}{4},\dfrac{{ - 1}}{3},\dfrac{{ - 5}}{{12}}) $
Now, the distance between two points is given by the formula,
$ d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
So the distance between points P and Q is
$
PQ = \sqrt {{{(\dfrac{{ - 1}}{4} - \dfrac{3}{2})}^2} + {{(\dfrac{{ - 1}}{3} - 2)}^2} + {{(\dfrac{{ - 5}}{{12}} - \dfrac{5}{2})}^2}} $
$= \sqrt {\dfrac{{49}}{{16}} + \dfrac{{49}}{9} + \dfrac{{1225}}{{144}}}$
$ = \sqrt {\dfrac{{441 + 784 + 1225}}{{144}}} = \sqrt {\dfrac{{2450}}{{144}}}$
$ \Rightarrow PQ = \dfrac{{35\sqrt 2 }}{{12}} \;
  $
Hence, the distance between the points P and Q is $ \dfrac{{35\sqrt 2 }}{{12}}units $ .
So, the correct answer is “$ \dfrac{{35\sqrt 2 }}{{12}}units $”.

Note: This question can also be solved by another method. From the equation of the planes we see that the planes are parallel, we can find out the distance between the planes and then by doing the dot product of the normal to one plane and the line cutting it, we can find the angle between the normal and the line. This way we can find the length of PQ.