Question

Show that the lines `(x - 3)/3 = (y - 3)/(-1), z - 1` = 0 and `(x - 6)/2 = (z - 1)/3, y - 2` = 0 intersect. Aslo find the point of intersection

Answer 1

`(x - 3)/3 = (y - 3)/(-1), z - 1` = 0 ⇒  z = 1

`(x - 6)/2 = (z - 1)/3, y - 2` = 0 ⇒ y = 2

(x₁, y₁, z₁) = (3, 3, 1) and (x₂, y₂, z2) = (6, 2, 1)

(b₁, b₂, b₃) = (3, –1, 0) and (d₁, d₂, d₃) = (2, 0, 3)

Condition for intersection of two lines

`|(x_2 - x_1, y_2 - y_1, z_2 - z_1),("b"_1, "b"_2, "b"_3),("d"_1, "d"_2, "d"_3)|` = 0

`|(3, -1, 0),(3, 1, 0),(2, 1, 3)|` = 0  Since (R₁ = R₂)

∴ Given two lines are intersecting lines.

Any point on the first time

`(x - 3)/3 = (y - 3)/(-1) = lambda` and z = 1

`(3lambda + 3, -lambda + 3, 1)`

Any point on the Second line

`(x - 6)/2 = (z - 1)/3 = mu` and y = 2

`(2mu + 6, 2, 3mu + 1)`

∴ `3mu + 1` = 1

`3mu` = 0

`mu` = 0

`-lambda + 3` = 2

`-lambda` = – 1

`lamda` = 1

∴ The required point of intersection is (6, 2, 1)