📝 Unit 8: Lines and Planes — Unit Test

All four equation forms · Distance · Configurations
✅ Graded
⏱️ 75 min  |  Total: /60 marks
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15
Part A: K/U [15]
1 [3]
Write vector and parametric equations of the line through \((2,-1,4)\) parallel to \(\vec d=(3,1,-2)\).
2 [3]
Write the symmetric equations of the line through \(A(1,2,3)\) and \(B(4,0,1)\).
3 [3]
Find a scalar equation of the plane through \((2,3,-1)\) with normal \((1,-2,3)\).
4 [3]
Find a scalar equation of the plane containing the three points \(P(1,0,0),\,Q(0,2,0),\,R(0,0,3)\).
5 [3]
MC: Two planes \(2x-3y+z=4\) and \(-4x+6y-2z=10\) are:
Part B: Thinking [15]
6 [5]
Find the distance from point \(P(2,3,1)\) to the plane \(2x-y+2z=10\).
7 [5]
Find the angle (in degrees) between the planes \(2x+y-z=5\) and \(x-y+3z=2\).
8 [5]
Find a scalar equation of the plane containing the line \((x,y,z)=(1,0,2)+t(2,-1,1)\) and the point \(P(3,1,0)\).
Part C: Communication [15]
9 [4]
Compare the four forms of a line equation in R³ (vector, parametric, symmetric, and the impossibility of a single scalar). When is each form most useful?
10 [4]
Describe step-by-step how to find the equation of a plane given (i) three points; (ii) a point and two direction vectors; (iii) a line and a point not on it.
11 [4]
Justify the formula for distance from a point to a plane using projection onto the normal.
12 [3]
Explain when a line in R³ is parallel to a plane (and not contained in it). Provide a test using direction and normal vectors.
Part D: Application [15]
13 [5]
A drone flies in a straight line through \((0,0,5)\) at velocity \((3,4,0)\) m/s. Find a parametric equation of its path. After how many seconds does the drone reach the plane \(x+y+z=20\)?
14 [5]
A satellite ground track lies in the plane \(2x+y-2z=8\) (km, with origin at Earth center, scaled). Find the closest point on the plane to the origin and its distance.
15 [5]
A robot arm endpoint moves along the line \(\vec r=(1,2,3)+t(2,-1,2)\). The work surface is the plane \(2x+3y-z=10\). Determine if the line meets the plane, and if so, the point of intersection.

Evaluation Rubric

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

1. Vector: \((x,y,z)=(2,-1,4)+t(3,1,-2)\). Parametric: \(x=2+3t,\,y=-1+t,\,z=4-2t\).

2. \(\vec{AB}=(3,-2,-2)\). \(\frac{x-1}{3}=\frac{y-2}{-2}=\frac{z-3}{-2}\).

3. \(1(x-2)-2(y-3)+3(z+1)=0\) → \(x-2y+3z+7=0\).

4. \(\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=1\) or \(6x+3y+2z=6\).

5. Normals \((2,-3,1)\) and \((-4,6,-2)=-2(2,-3,1)\) parallel; check \(P=(2,0,0)\) in first: \(4=4\) ✓; in second: \(-8+0+0=-8\ne 10\) → distinct parallel → (b).

6. \(|2(2)-3+2(1)-10|/\sqrt{4+1+4}=|{-7}|/3=7/3\approx 2.33\).

7. \(\vec n_1\cdot\vec n_2=2-1-3=-2\); \(|\vec n_1|=\sqrt 6,\,|\vec n_2|=\sqrt{11}\); \(\cos\theta=|-2|/\sqrt{66}\); \(\theta\approx 75.8°\).

8. Direction along line: \(\vec d=(2,-1,1)\); \(\vec{AP}\) from \((1,0,2)\) to \((3,1,0)\): \((2,1,-2)\). Normal \(=\vec d\times\vec{AP}=(-1)(-2)-(1)(1),\,(1)(2)-(2)(-2),\,(2)(1)-(-1)(2))=(1,6,4)\). Plane: \(1(x-1)+6(y-0)+4(z-2)=0\) → \(x+6y+4z-9=0\).

9. Vector: includes direction (geometric). Parametric: best for finding points/intersections. Symmetric: explicit relations. No single scalar in R³ because the solution set of one linear equation is a 2-D plane, not a 1-D line.

10. (i) \(\vec n=\vec{AB}\times\vec{AC}\); (ii) \(\vec n=\vec u\times\vec v\); (iii) get a 2nd direction from line point to external point, then \(\vec n=\vec d\times\vec{PA}\).

11. Vector from any plane point \(P_0\) to \(P\): \(\vec v\). Distance = \(|\text{proj}_{\vec n}\vec v|=|\vec v\cdot\vec n|/|\vec n|=|ax_0+by_0+cz_0-d|/\sqrt{a^2+b^2+c^2}\).

12. Line parallel to plane iff direction perpendicular to normal: \(\vec d\cdot\vec n=0\). Not contained iff a line point doesn't satisfy the plane equation.

13. \((x,y,z)=(0,0,5)+t(3,4,0)\). Sub: \(3t+4t+5=20\) → \(t=15/7\approx 2.14\) s.

14. Distance from origin = \(|0+0-0-8|/\sqrt{4+1+4}=8/3\). Closest point: along normal scaled: \(\frac{8}{9}(2,1,-2)=(16/9,8/9,-16/9)\).

15. Sub line into plane: \(2(1+2t)+3(2-t)-(3+2t)=10\) → \(2+4t+6-3t-3-2t=10\) → \(5-t=10\) → \(t=-5\). Point: \((1+2(-5),2-(-5),3+2(-5))=(-9,7,-7)\). Yes, intersects at \((-9,7,-7)\).