Question

If the three planes $x = 5,\;{\text{2x - 5ay + 3z - 8 = 0 and 3bx + y - 3z = 0}}$ contains a common line then (a, b) lies equal to
\[
  A.{\text{ (}}\dfrac{{ - 1}}{5}{\text{ }}\dfrac{8}{{15}}) \\
  B.{\text{ (}}\dfrac{1}{5}{\text{ }}\dfrac{{ - 8}}{{15}}) \\
  C.\;{\text{(}}\dfrac{{ - 8}}{{15}}\,\dfrac{1}{5}) \\
  D.{\text{ (}}\dfrac{8}{{15}}\dfrac{{ - 1}}{5}) \\
 \]

Answer 1

Hint: Here, we are given the value of one unknown term, place the given value in the equations. it, Using the two different equations and comparing it, will find the required points (a, b) which lie on the line.

Complete step by step answer:
 Three-planes are given –
$
  x = 5 \\
  2x - 5ay + 3z - 8 = 0 \\
  3bx + y - 3z = 0 \\
 $
Substitute $x = 5$ in the remaining two equations-
$
  2x - 5ay + 3z - 8 = 0 \\
  2(5) - 5ay + 3z - 8 = 0 \\
  10 - 5ay + 3z - 8 = 0 \\
   - 5ay + 3z + 2 = 0{\text{ }}.......{\text{(1)}} \\
 $ (Simplify using the basic mathematical operations)
$
  3bx + y - 3z = 0 \\
  3b(5) + y - 3z = 0 \\
  15b + y - 3z = 0{\text{ }}.......{\text{(2)}} \\
 $ (Simplify using the basic mathematical operations) Since equation (1) & equation (2) contains the common line – Comparing both the lines implies –
$
   - 5a = 1 \\
  \therefore a = \dfrac{{ - 1}}{5} \\
 $
And
$
  15b = 2 \\
  \therefore b = \dfrac{2}{{15}} \\
 $
$(a,b) = \left( {\dfrac{{ - 1}}{5},\dfrac{2}{{15}}} \right)$

Hence, from the given multiple choice, no option is the correct answer.

Note: Always remember the comparative properties of the different linear equations and lines. The systems of linear equations in two variables are of three types.
$1)$ An independent system – It has exactly one solution pair, the point at where the two lines intersect each other.
$2)$ An inconsistent system – It has no solution, as the two parallel lines never intersect each other.
$3)$ A dependent system – It has infinitely many solutions, as the lines coincident to each other.