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Question: The lines $\frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1}$ ...

The lines x−12=y+12=z−14\frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4} and x−31=y−62=z1\frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1} intersect each other at point

A

(-2, -4,5)

B

(-2,-4,-5)

C

(2,4,-5)

D

(2,-5)

Answer

(-2,-4,-5)

Explanation

Solution

To find the intersection point of two lines, we first parameterize each line.

The first line is given by: L1:x−12=y+12=z−14L_1: \frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4} Let this common ratio be λ\lambda. Then, any point on L1L_1 can be expressed as: x=2λ+1x = 2\lambda + 1 y=2λ−1y = 2\lambda - 1 z=4λ+1z = 4\lambda + 1

The second line is given by: L2:x−31=y−62=z1L_2: \frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1} Let this common ratio be μ\mu. Then, any point on L2L_2 can be expressed as: x=μ+3x = \mu + 3 y=2μ+6y = 2\mu + 6 z=μz = \mu

If the lines intersect, there must be a point (x,y,z)(x, y, z) that lies on both lines. Therefore, we equate the corresponding coordinates:

  1. 2λ+1=μ+32\lambda + 1 = \mu + 3
  2. 2λ−1=2μ+62\lambda - 1 = 2\mu + 6
  3. 4λ+1=μ4\lambda + 1 = \mu

From equation (3), we have μ=4λ+1\mu = 4\lambda + 1. Substitute this expression for μ\mu into equation (1): 2λ+1=(4λ+1)+32\lambda + 1 = (4\lambda + 1) + 3 2λ+1=4λ+42\lambda + 1 = 4\lambda + 4 1−4=4λ−2λ1 - 4 = 4\lambda - 2\lambda −3=2λ-3 = 2\lambda λ=−32\lambda = -\frac{3}{2}

Now, substitute the value of λ\lambda back into the expression for μ\mu: μ=4(−32)+1\mu = 4\left(-\frac{3}{2}\right) + 1 μ=−6+1\mu = -6 + 1 μ=−5\mu = -5

To confirm that the lines intersect, we must check if these values of λ\lambda and μ\mu satisfy the remaining equation (2): Substitute λ=−32\lambda = -\frac{3}{2} and μ=−5\mu = -5 into equation (2): 2(−32)−1=2(−5)+62\left(-\frac{3}{2}\right) - 1 = 2(-5) + 6 −3−1=−10+6-3 - 1 = -10 + 6 −4=−4-4 = -4 Since the equation holds true, the lines intersect.

Finally, substitute the value of λ\lambda (or μ\mu) into the parametric equations to find the intersection point. Using λ=−32\lambda = -\frac{3}{2} for L1L_1: x=2(−32)+1=−3+1=−2x = 2\left(-\frac{3}{2}\right) + 1 = -3 + 1 = -2 y=2(−32)−1=−3−1=−4y = 2\left(-\frac{3}{2}\right) - 1 = -3 - 1 = -4 z=4(−32)+1=−6+1=−5z = 4\left(-\frac{3}{2}\right) + 1 = -6 + 1 = -5

The intersection point is (−2,−4,−5)(-2, -4, -5).