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Question: Let $L_1$: $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $L_2$: $\frac{x-2}{3} = \frac{y-4}{4}...

Let L1L_1: x−12=y−23=z−34\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} and L2L_2: x−23=y−44=z−55\frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5} be two lines. Then which of the following points lies on the line of the shortest distance between L1L_1 and L2L_2?

A

(−53,−7,1)(\frac{-5}{3},-7,1)

B

(2,3,13)(2,3,\frac{1}{3})

C

(83,−1,13)(\frac{8}{3},-1,\frac{1}{3})

D

(143,−3,223)(\frac{14}{3},-3,\frac{22}{3})

Answer

Option (4) (143,−3,223)(\frac{14}{3}, -3, \frac{22}{3})

Explanation

Solution

To find the point on the line of shortest distance between L1L_1 and L2L_2, we need to:

  1. Parameterize the lines:

    L1:P(1+2t,2+3t,3+4t)L_1: P(1+2t, 2+3t, 3+4t)

    L2:Q(2+3s,4+4s,5+5s)L_2: Q(2+3s, 4+4s, 5+5s)

  2. Find the vector PQ:

    PQ=Q−P=(1+3s−2t,2+4s−3t,2+5s−4t)PQ = Q - P = (1+3s-2t, 2+4s-3t, 2+5s-4t)

  3. Apply perpendicularity conditions:

    PQ⋅a⃗=0PQ \cdot \vec{a} = 0 and PQ⋅b⃗=0PQ \cdot \vec{b} = 0, where a⃗=(2,3,4)\vec{a} = (2,3,4) and b⃗=(3,4,5)\vec{b} = (3,4,5) are direction vectors of L1L_1 and L2L_2 respectively.

    This gives us two equations:

    2(1+3s−2t)+3(2+4s−3t)+4(2+5s−4t)=0⇒16+38s−29t=02(1+3s-2t) + 3(2+4s-3t) + 4(2+5s-4t) = 0 \Rightarrow 16 + 38s - 29t = 0

    3(1+3s−2t)+4(2+4s−3t)+5(2+5s−4t)=0⇒21+50s−38t=03(1+3s-2t) + 4(2+4s-3t) + 5(2+5s-4t) = 0 \Rightarrow 21 + 50s - 38t = 0

  4. Solve for s and t:

    Solving the system of equations, we get s=−16s = -\frac{1}{6} and t=13t = \frac{1}{3}.

  5. Find points P and Q:

    P=(1+23,2+1,3+43)=(53,3,133)P = (1+\frac{2}{3}, 2+1, 3+\frac{4}{3}) = (\frac{5}{3}, 3, \frac{13}{3})

    Q=(2−12,4−23,5−56)=(32,103,256)Q = (2-\frac{1}{2}, 4-\frac{2}{3}, 5-\frac{5}{6}) = (\frac{3}{2}, \frac{10}{3}, \frac{25}{6})

  6. Determine the direction vector of the shortest distance line:

    Q−P=(32−53,103−3,256−133)=(−16,13,−16)=16(−1,2,−1)Q - P = (\frac{3}{2}-\frac{5}{3}, \frac{10}{3}-3, \frac{25}{6}-\frac{13}{3}) = (-\frac{1}{6}, \frac{1}{3}, -\frac{1}{6}) = \frac{1}{6}(-1, 2, -1)

  7. Parameterize the shortest distance line using point P:

    R(t)=(53,3,133)+t(−1,2,−1)R(t) = (\frac{5}{3}, 3, \frac{13}{3}) + t(-1, 2, -1)

  8. Check the options:

    For option (4): (143,−3,223)(\frac{14}{3}, -3, \frac{22}{3})

    53−t=143⇒t=−3\frac{5}{3} - t = \frac{14}{3} \Rightarrow t = -3

    Check y: 3+2(−3)=−33 + 2(-3) = -3

    Check z: 133−(−3)=223\frac{13}{3} - (-3) = \frac{22}{3}

    Since option (4) satisfies the parameterized equation, it lies on the line of shortest distance.