Pairs of Lines in 3D (Edexcel A-Level Further Mathematics): Revision Notes

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6.1.2 Pairs of Lines in 3D

Parallel and Skew Lines

  1. Parallel Lines: Two lines are parallel if their direction vectors are scalar multiples:
b1=kb2,k∈R\mathbf{b}_1 = k \mathbf{b}_2, \quad k \in \mathbb{R}
  1. Skew Lines: Skew lines are not parallel and do not intersect. They exist in 3D space without lying in the same plane.

Finding the Intersection of Two Lines

Two lines intersect if there exists a common point.

Let:

r1=a1+λb1,r2=a2+μb2\mathbf{r}_1 = \mathbf{a}_1 + \lambda \mathbf{b}_1, \quad \mathbf{r}_2 = \mathbf{a}_2 + \mu \mathbf{b}_2

At the intersection, r1=r2\mathbf{r}_1 = \mathbf{r}_2, giving three equations:

a1i+λb1i=a2i+μb2i,for i=1,2,3a_{1i} + \lambda b_{1i} = a_{2i} + \mu b_{2i}, \quad \text{for } i = 1, 2, 3

Solve these simultaneously to find λ\lambda and μ\mu.

If a consistent solution exists, substitute λ\lambda or μ\mu back to find the intersection point.

Worked Example

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Example: Find the Intersection of Two Lines

Find the intersection of:

r1=(123)+λ(1−12),r2=(201)+μ(−213)\mathbf{r}_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}, \quad \mathbf{r}_2 = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix}

Step 1: Set equations equal:

(1+λ2−λ3+2λ)=(2−2μμ1+3μ)\begin{pmatrix} 1 + \lambda \\ 2 - \lambda \\ 3 + 2\lambda \end{pmatrix} = \begin{pmatrix} 2 - 2\mu \\ \mu \\ 1 + 3\mu \end{pmatrix}

Step 2**: Write component equations:**

  • 1+λ=2−2μ1 + \lambda = 2 - 2\mu
  • 2−λ=μ2 - \lambda = \mu
  • 3+2λ=1+3μ3 + 2\lambda = 1 + 3\mu

Step 3**: Solve simultaneously:**

From: 2−λ=μ:μ=2−λ2 - \lambda = \mu: \mu = 2 - \lambda

Substitute μ=2−λ\mu = 2 - \lambda into 1+λ=2−2μ1 + \lambda = 2 - 2\mu:

1+λ=2−2(2−λ)1 + \lambda = 2 - 2(2-\lambda)

Simplify:

1+λ=2−4+2λ  ⟹  1=−2+λ  ⟹  λ=31 + \lambda = 2 - 4 + 2\lambda \implies 1 = -2 + \lambda \implies \lambda = 3

Substitute λ=3\lambda = 3 into μ=2−λ\mu = 2 - \lambda

μ=2−3=−1.\mu = 2 - 3 = -1.

Step 4**: Find the intersection point:**

Substitute λ=3\lambda = 3 into r1\mathbf{r}_1:

r1=(123)+3(1−12)=(4−19)\mathbf{r}_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + 3 \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \\ 9 \end{pmatrix}

Result:

The lines intersect at (4, -1, 9)

Note Summary

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Common Mistakes:

  1. Confusing the definitions of parallel and skew lines: Parallel lines have direction vectors that are scalar multiples, while skew lines do not intersect and are not parallel.

  2. Errors in simultaneous equations when finding intersections: Mismanagement of algebra when solving λ\lambda and μ\mu can lead to incorrect results.

  3. Assuming lines must intersect: Always check the solution for consistency; lines may be skew and not intersect.

  4. Misinterpreting vector components: Ensure proper alignment of vector components when forming equations from vector representations of lines.

  5. Omitting back-substitution to confirm intersection points: Forgetting to substitute λ\lambda or μ\mu into the original equations may result in unverified or incorrect solutions.

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Key Formulas:

  1. Parallel Lines: Two lines r1=a1+λb1\mathbf{r}_1 = \mathbf{a}_1 + \lambda \mathbf{b}_1 and r2=a2+μb2\mathbf{r}_2 = \mathbf{a}_2 + \mu \mathbf{b}_2 are parallel if:
b1=kb2,k∈R\mathbf{b}_1 = k \mathbf{b}_2, \quad k \in \mathbb{R}
  1. Intersection of Two Lines: Lines intersect if:
a1+λb1=a2+μb2\mathbf{a}_1 + \lambda \mathbf{b}_1 = \mathbf{a}_2 + \mu \mathbf{b}_2

Solve for λ\lambda and μ\mu by forming three simultaneous equations for the x,yx, y, and z z components.

  1. Skew Lines: Skew lines do not satisfy the conditions for parallelism or intersection.

  2. Intersection Point Verification: After finding λ\lambda and μ\mu, substitute back into r1\mathbf{r}_1 or r2\mathbf{r}_2 to determine the intersection point.

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