Show that the lines 5-x/-4 = y-7/4 = z+3/-5 and x-8/7 = 2y-8/2 = z-5/3 are coplanar.

Question

Show that the lines \(\frac{5 - x}{-4} = \frac{y -7}{4} = \frac{z + 3}{-5}\) and \(\frac{x -8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}\) are coplanar.

Answer 1

Given lines can be written as

\(\frac{x - 5}{4} = \frac{y -7}{4} = \frac{z + 3}{-5}\)

and \(\frac{x -8}{7} = \frac{y - 4}{1} = \frac{z - 5}{3}\)

On comparing both lines with,

\(\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{b}\)

respectively, we get

x₁ = 5, y₁ = 7, z₁ = - 3, a₁ = 4, b₁ = 4, c₁ = -5

and x₂ = 8, y₂ = 4, z₂ = 5, a₂ = 7, b₂ = 1, c₂ = 3

If given lines are coplanar, then

= 3(12 + 5) + 3(12 + 35) + 8(4 - 28)

= 3 × 17 + 3 × 47 + 8(- 24)

= 51 + 141 - 192

= 192 - 192 = 0 = RHS

Therefore, given lines are coplanar.

Hence proved.