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More from Three Dimensional Geometry
MathMedAdv 2019 • Paper 2

Find Points on Line for Collinear Intersections Across Three Skew Lines

Three lines are given as follows:

L1:r⃗=λi^,λ∈R,L2:r⃗=k^+μj^,μ∈R,andL3:r⃗=i^+j^+νk^,ν∈R\begin{array}{ll} L_1: & \vec{r} = \lambda \hat{i}, \quad \lambda \in \mathbb{R}, \\ L_2: & \vec{r} = \hat{k} + \mu \hat{j}, \quad \mu \in \mathbb{R}, \quad \text{and} \\ L_3: & \vec{r} = \hat{i} + \hat{j} + \nu \hat{k}, \quad \nu \in \mathbb{R} \end{array}

For which point(s) QQ on L2L_2 is it possible to find a point PP on L1L_1 and a point RR on L3L_3 such that P,Q,P, Q, and RR are collinear?

Options

A

k^−12j^\hat{k} - \frac{1}{2} \hat{j}

Correct
B

k^\hat{k}

C

k^+12j^\hat{k} + \frac{1}{2} \hat{j}

Correct
D

k^+j^\hat{k} + \hat{j}

Topics & Concepts

Three Dimensional GeometryLine in 3DDot and Cross Product

Step-by-Step Solution

To determine the points QQ on the line L2L_2 for which there exist points PP on L1L_1 and RR on L3L_3 such that P,Q,P, Q, and RR are collinear, let us represent the coordinates of the general points on each line:

  1. A point PP on L1L_1 can be written as: P=(λ,0,0),λ∈RP = (\lambda, 0, 0), \quad \lambda \in \mathbb{R}

  2. A point QQ on L2L_2 can be written as: Q=(0,μ,1),μ∈RQ = (0, \mu, 1), \quad \mu \in \mathbb{R}

  3. A point RR on L3L_3 can be written as: R=(1,1,ν),ν∈RR = (1, 1, \nu), \quad \nu \in \mathbb{R}

For the three points P,Q,P, Q, and RR to be collinear, the vectors PQ⃗\vec{PQ} and QR⃗\vec{QR} must be collinear (parallel).

We have: PQ⃗=Q−P=(−λ)i^+μj^+k^\vec{PQ} = Q - P = (-\lambda)\hat{i} + \mu\hat{j} + \hat{k} QR⃗=R−Q=i^+(1−μ)j^+(ν−1)k^\vec{QR} = R - Q = \hat{i} + (1 - \mu)\hat{j} + (\nu - 1)\hat{k}

For PQ⃗\vec{PQ} and QR⃗\vec{QR} to be parallel, their corresponding directional ratios must be proportional: −λ1=μ1−μ=1ν−1\frac{-\lambda}{1} = \frac{\mu}{1 - \mu} = \frac{1}{\nu - 1}

Let us analyze the conditions under which these ratios are valid:

  • If μ=0\mu = 0, the yy-component of PQ⃗\vec{PQ} is 00 while the yy-component of QR⃗\vec{QR} is 11. For the vectors to be parallel, either P=QP = Q or PQ⃗=0⃗\vec{PQ} = \vec{0}, which is impossible since the zz-coordinate of PP is 00 and that of QQ is 11. Thus, μ≠0\mu \neq 0.
  • If μ=1\mu = 1, the yy-component of QR⃗\vec{QR} is 00 while the yy-component of PQ⃗\vec{PQ} is 11. Similarly, this requires Q=RQ = R, which is impossible as their xx-coordinates differ (0≠10 \neq 1). Thus, μ≠1\mu \neq 1.

For any μ∈R∖{0,1}\mu \in \mathbb{R} \setminus \{0, 1\}, the middle term is well-defined and non-zero. We can then uniquely solve for λ\lambda and ν\nu: λ=μμ−1∈R\lambda = \frac{\mu}{\mu - 1} \in \mathbb{R} ν=1+1−μμ=1μ∈R\nu = 1 + \frac{1 - \mu}{\mu} = \frac{1}{\mu} \in \mathbb{R}

Thus, collinear points P,Q,RP, Q, R can be found if and only if μ≠0\mu \neq 0 and μ≠1\mu \neq 1.

Now, let us test the given options for Q=k^+μj^Q = \hat{k} + \mu \hat{j}:

  • (A) k^−12j^  ⟹  μ=−12\hat{k} - \frac{1}{2}\hat{j} \implies \mu = -\frac{1}{2} (≠0,1\neq 0, 1)   ⟹  \implies Possible
  • (B) k^  ⟹  μ=0\hat{k} \implies \mu = 0   ⟹  \implies Not possible
  • (C) k^+12j^  ⟹  μ=12\hat{k} + \frac{1}{2}\hat{j} \implies \mu = \frac{1}{2} (≠0,1\neq 0, 1)   ⟹  \implies Possible
  • (D) k^+j^  ⟹  μ=1\hat{k} + \hat{j} \implies \mu = 1   ⟹  \implies Not possible

Therefore, the correct options are A and C.

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