Porisma

Worked solution

Distinguish parallel, intersecting and skew lines algebraically

For each pair, test whether the lines are parallel, intersecting or skew. If they intersect, find the position vector of their intersection. All parameters are real.

(a) l1:r=(1,2,3)+λ(1,2,1)l_1:\mathbf r=(1,2,3)+\lambda(1,2,1) and l2:r=(1,3,4)+μ(2,4,2)l_2:\mathbf r=(1,3,4)+\mu(2,4,2).

(b) l1:r=(1,2,3)+λ(1,2,1)l_1:\mathbf r=(1,2,3)+\lambda(1,2,1) and l2:r=(0,1,5)+μ(1,3,4)l_2:\mathbf r=(0,1,5)+\mu(1,3,4).

(c) l1:r=(1,2,3)+λ(1,2,1)l_1:\mathbf r=(1,2,3)+\lambda(1,2,1) and l2:r=(0,1,2)+μ(1,3,4)l_2:\mathbf r=(0,1,2)+\mu(1,3,4).

CJC · Lecture · Example 25

(a) Compare direction vectors

(2,4,2)=2(1,2,1)(2,4,2)=2(1,2,1), so the directions are parallel. The source concludes parallel lines. To distinguish them from coincident lines, the base-point displacement (1,3,4)−(1,2,3)=(0,1,1)(1,3,4)-(1,2,3)=(0,1,1) is not a multiple of (1,2,1)(1,2,1); thus these are distinct parallel lines.

How did this part go?

(b) Method 1: solve and check the remaining coordinate

The directions (1,2,1)(1,2,1) and (1,3,4)(1,3,4) are not scalar multiples. Assuming intersection gives 1+λ=μ1+\lambda=\mu, 2+2λ=1+3μ2+2\lambda=1+3\mu, 3+λ=5+4μ3+\lambda=5+4\mu, or (1) λ−μ=−1\lambda-\mu=-1, (2) 2λ−3μ=−12\lambda-3\mu=-1, (3) λ−4μ=2\lambda-4\mu=2. Subtracting (3) from (1) gives 3μ=−33\mu=-3, so μ=−1; (1) then gives λ=−2. In (2), 2(−2)−3(−1)=−12(-2)-3(-1)=-1 equals the right side. All three equations are satisfied: the lines intersect.

How did this part go?

(b) Method 2: retain the calculator system and read the intersection

The source G.C. PlySmlt 2 screenshot enters the augmented system with columns λ,μ, right side:

[1−1−12−3−11−42].\left[\begin{array}{rr|r}1&-1&-1\\2&-3&-1\\1&-4&2\end{array}\right].

Its displayed solution is X1=−2, X2=−1, corresponding to λ=−2 and μ=−1. Therefore the intersection position vector is (1,2,3)−2(1,2,1)=(−1,−2,1)(1,2,3)-2(1,2,1)=(-1,-2,1); substituting μ=−1 into the other line gives the same point.

How did this part go?

(c) Method 1: expose the inconsistent coordinate

The directions are again nonparallel. Equating components gives (1) λ−μ=−1\lambda-\mu=-1, (2) 2λ−3μ=−12\lambda-3\mu=-1, (3) λ−4μ=−1\lambda-4\mu=-1. Subtract (3) from (1): 3μ=03\mu=0, so μ=0 and λ=−1. But in (2), 2(−1)−3(0)=−2≠−12(-1)-3(0)=-2\ne-1. No parameter pair satisfies all three equations; the nonparallel lines are skew.

How did this part go?

(c) Method 2: retain the inconsistent calculator system

The second source G.C. screenshot enters

[1−1−12−3−11−4−1].\left[\begin{array}{rr|r}1&-1&-1\\2&-3&-1\\1&-4&-1\end{array}\right].

Its output is NO SOLUTION FOUND. This is consistent with the unused-coordinate contradiction. Since parallel directions were already ruled out, no intersection means the lines are skew, not parallel.

How did this part go?

Common mistakes

Use distinct parameters for the two lines. A solution of only two component equations is not enough. Rule out parallel directions before concluding that nonintersection means skew.

Solution notes

Both analytic and calculator methods, all augmented-matrix entries and both printed calculator outputs retained. The calculator figures are represented as accessible equations/output, not raster screenshots. The coincidence check in(a) is supplementary to the source parallel conclusion.

Read the hint