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Worked solution

Classify four pairs of spatial lines and find acute angles

Find whether the following pairs of lines are parallel, intersecting or skew. If they intersect, find the point of intersection and the acute angle between the lines.

(a) r=(1,−1,1)+λ(3,4,1)\mathbf r=(1,-1,1)+\lambda(3,4,1), λ∈R\lambda\in\mathbb R, and r=(1,2,3)+μ(9,12,3)\mathbf r=(1,2,3)+\mu(9,12,3), μ∈R\mu\in\mathbb R.

(b) r=(1,0,3)+s(2,1,1)\mathbf r=(1,0,3)+s(2,1,1), s∈Rs\in\mathbb R, and r=(2,−1,1)+t(1,−2,0)\mathbf r=(2,-1,1)+t(1,-2,0), t∈Rt\in\mathbb R.

(c) r=4i+8j+3k+α(i+2j+k)\mathbf r=4\mathbf i+8\mathbf j+3\mathbf k+\alpha(\mathbf i+2\mathbf j+\mathbf k), α∈R\alpha\in\mathbb R, and r=7i+6j+5k+β(6i+4j+5k)\mathbf r=7\mathbf i+6\mathbf j+5\mathbf k+\beta(6\mathbf i+4\mathbf j+5\mathbf k), β∈R\beta\in\mathbb R.

(d) r=(4,2,1)+λ(2,1,−2)\mathbf r=(4,2,1)+\lambda(2,1,-2), λ∈R\lambda\in\mathbb R, and the z-axis.

[Hint: The z-axis has equation r=γ(0,0,1)\mathbf r=\gamma(0,0,1), γ∈R\gamma\in\mathbb R.]

CJC · Discussion · Question 5

Answer

(a) Parallel. (b) Skew. (c) (1,2,0)(1,2,0), 27.9∘27.9^\circ. (d) (0,0,5)(0,0,5), 48.2∘48.2^\circ.

(a) Parallel and distinct

(9,12,3)=3(3,4,1)(9,12,3)=3(3,4,1), so the directions are parallel. Reference-point difference (0,3,2)(0,3,2) is not a multiple of (3,4,1)(3,4,1), confirming distinct parallel lines.

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(b) Check consistency: skew

The directions are not scalar multiples. Intersection would require

2s−t=1,s+2t=−1,s=−2.2 s-t=1,\quad s+2 t=-1,\quad s=-2.

The last two give s=−2s=-2, t=1/2t=1/2. The first left side is then −9/2-9/2, not 1. Non-parallel lines without intersection are skew.

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(c) Find the intersection

The equations are

α−6β=3,2α−4β=−2,α−5β=2.\alpha-6\beta=3,\quad 2\alpha-4\beta=-2,\quad\alpha-5\beta=2.

The first two give α=−3\alpha=-3, β=−1\beta=-1; these also satisfy the third. Substitution gives (1,2,0)(1,2,0).

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(c) Acute angle

The direction dot product is 19 and squared norms are 6 and 77. Thus

θ=cos⁡−1(19677)≈27.9∘\theta=\cos^{-1}\left(\frac{19}{\sqrt 6\sqrt{77}}\right)\approx 27.9^\circ

to 1 d.p., or 0.487 rad to 3 s.f.

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(d) Intersect the z-axis

Solve 4+2λ=04+2\lambda=0, 2+λ=02+\lambda=0, 1−2λ=γ1-2\lambda=\gamma. Both first equations require λ=−2\lambda=-2, giving γ=5\gamma=5. The point is (0,0,5)(0,0,5).

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(d) Select the acute supplement

The direction-vector angle is cos⁡−1(−2/3)=131.810…∘\cos^{-1}(-2/3)=131.810\ldots{}^\circ, or 2.3005…rad. The acute line angle is its supplement: 48.2∘48.2^\circ (1 d.p.), or 0.841 rad (3 s.f.). Equivalently use cos⁡−1(∣−2∣/3)\cos^{-1}(|-2|/3).

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Common mistakes

Non-parallel 3 Dlines need not intersect. Take the acute supplement when the direction-vector angle is obtuse.

Solution notes

All source cases and consistency equations retained, including degree and radian results. No source diagram is printed here.

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