📝 Unit 9: Intersections — Unit Test

Line–Line · Line–Plane · Plane–Plane · Three Planes
✅ Graded
⏱️ 75 min  |  Total: /60 marks
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15
Part A: K/U [15]
1 [3]
Find the point of intersection of \(L_1: (1,2)+t(2,3)\) and \(L_2: (5,1)+s(-1,2)\) (in R²).
2 [3]
Determine whether \(L_1: (1,0,1)+t(2,1,-1)\) and \(L_2: (3,1,0)+s(2,1,-1)\) are parallel, coincident, intersecting, or skew.
3 [3]
Find the intersection of line \((x,y,z)=(2,-1,3)+t(1,2,-1)\) with plane \(x+y+2z=8\).
4 [3]
Find a vector equation of the line of intersection of \(\pi_1: x+2y-z=4\) and \(\pi_2: 2x-y+z=3\).
5 [3]
MC: Three planes meet in a single common point. The augmented matrix has:
Part B: Thinking [15]
6 [5]
Solve: \(x+y+z=6,\ 2x-y+3z=14,\ x+4y-z=-2\). Determine if there's a unique solution; if so, find it.
7 [5]
Show that \(L_1: (0,1,2)+t(1,1,1)\) and \(L_2: (1,0,3)+s(2,3,1)\) are skew. Find the (perpendicular) distance between them. Use \(d=\frac{|\vec{P_1P_2}\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}\).
8 [5]
A system of three planes \(\pi_1:x+y+z=2\), \(\pi_2:2x+2y+2z=4\), \(\pi_3:x-y+z=0\). Determine if the system is consistent and describe the geometric configuration.
Part C: Communication [15]
9 [4]
Describe step-by-step procedures for each of: (a) two lines in R²; (b) two lines in R³; (c) a line and a plane; (d) two planes. Include the algebra each requires.
10 [4]
Catalogue all possible configurations of three planes in R³ (single point, line, coincident, no common point — and sub-cases). Sketch (in words) each.
11 [4]
Explain why two non-parallel planes always intersect in a line. Why is the direction of the line given by \(\vec n_1\times\vec n_2\)?
12 [3]
Justify why the formula for distance between skew lines uses the cross product of the two direction vectors.
Part D: Application [15]
13 [5]
Two airplanes fly along straight paths: \(A_1:(0,0,1)+t(2,3,0.1)\) and \(A_2:(5,0,1)+s(0,3,0)\) (km, with \(t,s\) in min). Determine if their paths intersect, and if so, will the planes collide (meet at the same time)?
14 [5]
A laser pointer fires from \((1,2,5)\) in direction \((2,3,-4)\). A flat target plate occupies the plane \(x+2y-z=10\). Find where the beam strikes.
15 [5]
Three drone flight corridors are defined by the planes \(2x+y-z=4\), \(x-y+z=2\), \(3x+2y-2z=6\). Determine if there is a common safe meeting point.

Evaluation Rubric

LevelDescription%
4Thorough80–100
3Considerable70–79
2Some60–69
1Limited50–59
RInsufficient<50
📕 Answer Key

1. \(1+2t=5-s\) and \(2+3t=1+2s\). From first: \(s=4-2t\); sub into second: \(2+3t=1+2(4-2t)=9-4t\) → \(7t=7\) → \(t=1\), \(s=2\). Point: \((3,5)\).

2. Same direction \((2,1,-1)\). Test if \((3,1,0)\) on \(L_1\): \(3=1+2t\) → \(t=1\); then \(y=0+1=1\) ✓; \(z=1-1=0\) ✓. → coincident.

3. \((2+t)+(-1+2t)+2(3-t)=8\) → \(2+t-1+2t+6-2t=8\) → \(7+t=8\) → \(t=1\). Point: \((3,1,2)\).

4. \(\vec d=\vec n_1\times\vec n_2=(1,2,-1)\times(2,-1,1)=(2\cdot 1-(-1)(-1),\,(-1)(2)-(1)(1),\,(1)(-1)-(2)(2))=(1,-3,-5)\). Set \(z=0\): \(x+2y=4,\ 2x-y=3\) → \(y=(2x-3)\), so \(x+2(2x-3)=4\) → \(5x=10\) → \(x=2,y=1\). Line: \((2,1,0)+t(1,-3,-5)\).

5. Unique point ↔ rank 3 → (b).

6. Add (1)+(3): \(2x+5y=4\). (1)·2 - (2): \(0+3y-z=...\) easier: from (1) \(z=6-x-y\); sub into (2): \(2x-y+3(6-x-y)=14\) → \(-x-4y=-4\) → \(x+4y=4\). System: \(2x+5y=4,\ x+4y=4\). \(2(x+4y)-( 2x+5y)=8-4\) → \(3y=4\) → \(y=4/3\), \(x=4-16/3=-4/3\), \(z=6-(-4/3)-4/3=6=18/3\). Solution \((-4/3,4/3,6)\). Verify (3): \(-4/3+16/3-6=12/3-6=4-6=-2\) ✓.

7. \(\vec d_1\times\vec d_2=(1,1,1)\times(2,3,1)=(1\cdot1-1\cdot3,\,1\cdot2-1\cdot1,\,1\cdot3-1\cdot2)=(-2,1,1)\). \(\vec{P_1P_2}=(1,-1,1)\). Dot: \(-2-1+1=-2\). \(|\vec d_1\times\vec d_2|=\sqrt{6}\). Distance \(=2/\sqrt 6\approx 0.816\). Skew because \((-2)\ne 0\).

8. \(\pi_2=2\pi_1\) (same plane). \(\pi_3\) intersects \(\pi_1\) in a line (different normal). System reduces to 2 effective equations in 3 unknowns → infinitely many solutions along that line.

9. (a) Set up 2 eqs in 2 params \(s,t\), solve. (b) Same, but check consistency in 3rd component (otherwise skew or parallel). (c) Substitute parametric line into plane scalar equation, solve for \(t\). (d) Solve 2-eq 3-unknown system: assign one variable as parameter.

10. Cases: (i) Single point — three planes mutually intersecting. (ii) Common line — two planes coincident plus one intersecting; or all three meeting along the line. (iii) Coincident — all three same. (iv) No common point — parallel planes (one or more), or "triangular prism" (each pair meets in a line, no common line).

11. Two non-parallel normals span the orthogonal complement direction. The intersection set satisfies both linear equations → a 1-parameter family (line). Direction perpendicular to both normals = \(\vec n_1\times\vec n_2\).

12. \(\vec d_1\times\vec d_2\) is perpendicular to both lines' directions. Project the displacement \(\vec{P_1P_2}\) onto this common-perpendicular direction → shortest distance.

13. Set positions equal: \(2t=5\), \(3t=3s\), \(1+0.1t=1\). From 1st: \(t=2.5\); from 3rd: \(t=0\). Contradiction → paths don't even cross at same place → no intersection of paths in 3D (skew). No collision.

14. \((1+2t)+2(2+3t)-(5-4t)=10\) → \(1+2t+4+6t-5+4t=10\) → \(12t=10\) → \(t=5/6\). Point: \((1+5/3,2+5/2,5-10/3)=(8/3,9/2,5/3)\).

15. Try eliminating: (1)+(2): \(3x+0+0=6\) → \(x=2\). Sub into (1): \(4+y-z=4\) → \(y=z\). Sub into (3): \(6+2y-2y=6\) ✓ identically. So infinitely many solutions: \((2,t,t)\). Three planes meet in a line — common safe corridor exists.