Question

Find the equation of the line which intersects the lines x+2/1=y-3/2=z+1/4

Answer 1

The complete question is:

Find the equation of the line which intersects the lines x+2/1=y-3/2=z+1/4 and x-1/2=y-2/3=z-3/4 and passes through the point (1, 1, 1)

Given,

Equation of a first line:

x+2/1 = y-3/2 = z+1/4

⇒ General point is:

(λ - 2, 2λ + 3, 4λ - 1)

Equation of a second line:

x-1/2 = y-2/3 = z-3/4

⇒ General point is:

(2μ + 1, 3μ + 2, 4μ + 3)

The direction ratios of the the intersecting with above lines is,

(λ - 3, 2λ + 2, 4λ - 2)

And, the direction ratios of the required line may be,

(2μ, 3μ + 2, 4μ - 2)

Therefore, we have,

λ - 3 / 2μ = 2λ + 2 / 3μ + 2 = 4λ - 2 / 4μ - 2

⇒ λ - 3 / 2μ = 2λ + 2 / 3μ + 2 = 2λ - 1 / 2μ - 1

Let us say,

λ - 3 / 2μ = 2λ + 2 / 3μ + 2 = 2λ - 1 / 2μ - 1  = k ..........(1)

Therefore, we get,

λ - 3 = 2μk, 2λ + 2 = (3μ + 2)k, 2λ - 1 = (2μ - 1)k

solving the equation by substitution method, we get,

k = [ 4λ + 4 - 3λ +9 ] / 2

using (1), we get,

= λ + 2, λ = 9

μ = 3/11

Therefore, the direction cosines of the required line are (3, 10, 17) or (6, 20, 34)

Hence the required equation of intersecting line is x-1/3 = y-1/10 = z-1/17