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

The lines x−13=y+12=z−13\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{3} and x+24=y−13=z+1−2\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}

A

Intersect each other and point of intersection is (3, 2, 5)

B

Intersect each other and point of intersection is (-2, -1, -1)

C

Intersect each other and point of intersection is (4, 3, -2)

D

do not intersect.

Answer

The lines do not intersect.

Explanation

Solution

To determine if the lines intersect, we convert them to parametric form, equate the x and y coordinates to find the parameters, and then check if the z coordinates are consistent.

  1. Parametric equations:

    • For the first line:

      x=1+3t,y=−1+2t,z=1+3tx = 1 + 3t, \quad y = -1 + 2t, \quad z = 1 + 3t

    • For the second line:

      x=−2+4s,y=1+3s,z=−1−2sx = -2 + 4s, \quad y = 1 + 3s, \quad z = -1 - 2s

  2. Equate the x-coordinates:

    1+3t=−2+4s  ⟹  t=−1+43s1 + 3t = -2 + 4s \implies t = -1 + \frac{4}{3}s

  3. Equate the y-coordinates:

    −1+2t=1+3s  ⟹  t=1+32s-1 + 2t = 1 + 3s \implies t = 1 + \frac{3}{2}s

  4. Compare the two expressions for tt:

    −1+43s=1+32s-1 + \frac{4}{3}s = 1 + \frac{3}{2}s

    Multiplying both sides by 6 to eliminate fractions:

    −6+8s=6+9s  ⟹  −12=s-6 + 8s = 6 + 9s \implies -12 = s

    Then,

    t=−1+43(−12)=−1−16=−17t = -1 + \frac{4}{3}(-12) = -1 - 16 = -17

  5. Check the z-coordinates:

    • For the first line:

      z=1+3(−17)=−50z = 1 + 3(-17) = -50

    • For the second line:

      z=−1−2(−12)=23z = -1 - 2(-12) = 23

    Since −50≠23-50 \neq 23, the lines do not intersect.

Therefore, the lines do not intersect.