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Grade 12 · Lesson 5.2

The equation of a line in space

Name: Date: Time spent: min

Goals

Remember

Exercises

Exercise 1

Write parametric equations of the line through A(2,−1,3)A(2, -1, 3) with direction vector u⃗=(1,4,−2)\vec{u}=(1, 4, -2). Find the point for t=−1t=-1 and check whether B(4,7,−1)B(4, 7, -1) lies on the line.

Exercise 2

Write parametric and Cartesian equations of the line through A(1,0,−2)A(1, 0, -2) and B(3,2,−1)B(3, 2, -1).

Exercise 3

Write the equation of the line dd through M(1,−1,2)M(1, -1, 2) perpendicular to (P):2x−y+3z+5=0(P): 2x-y+3z+5=0. Find the point where dd meets (P)(P).

Exercise 4

Write parametric and Cartesian equations of the line through M(0,1,−1)M(0, 1, -1) parallel to Δ:x−12=y+3−1=z3\Delta: \dfrac{x-1}{2}=\dfrac{y+3}{-1}=\dfrac{z}{3}.

Exercise 5

Let d1:{x=1+2ty=−tz=3+td_1: \begin{cases}x=1+2t\\ y=-t\\ z=3+t\end{cases} and d2:x−3−4=y+12=z−4−2d_2: \dfrac{x-3}{-4}=\dfrac{y+1}{2}=\dfrac{z-4}{-2}. Is each claim true or false?

a) u⃗1=(2,−1,1)\vec{u}_1=(2, -1, 1) is a direction vector of d1d_1.

b) The point M(3,−1,4)M(3, -1, 4) lies on d1d_1.

c) d1d_1 is parallel to d2d_2.

d) d1d_1 is perpendicular to the plane (P):2x−y+z=0(P): 2x-y+z=0.

Exercise 6

An aeroplane takes off from the point O(0,0,0)O(0, 0, 0) and flies in a straight line, moving by the vector v⃗=(80,60,10)\vec{v}=(80, 60, 10) every second (unit: metre). When it reaches a height of 15001500 m, how many kilometres from OO is the point on the ground directly below it?

Exercise 7

Let d:x=2+td: x=2+t, y=1−ty=1-t, z=3+2tz=3+2t. Decide the position of dd relative to (P1):x+y−1=0(P_1): x+y-1=0, (P2):x+3y+z−8=0(P_2): x+3y+z-8=0, (P3):x−y+z−6=0(P_3): x-y+z-6=0.

Exercise 8

Decide the position of the lines d1:x=1+td_1: x=1+t, y=1−ty=1-t, z=2tz=2t and d2:x=1+2sd_2: x=1+2s, y=−2+sy=-2+s, z=5−sz=5-s. If they intersect, find the point of intersection.

Exercise 9

(Revisiting Operations on vectors in coordinates.) In the previous lesson we proved that A(1,2,−1)A(1, 2, -1), B(3,1,1)B(3, 1, 1), C(7,−1,5)C(7, -1, 5) are collinear using parallel vectors. Write parametric equations of the line ABAB and give the value of tt for the point CC.

Exercise 10 Challenge

A laser beam starts at A(1,2,3)A(1, 2, 3), reflects off a flat mirror lying in the plane (P):x+2y+2z−2=0(P): x+2y+2z-2=0, and then reaches B(3,2,2)B(3, 2, 2). Find the point MM of the mirror where it reflects. (Hint: the path A→M→BA\to M\to B is shortest when MM lies on the segment A′BA'B, where A′A' is the reflection of AA in (P)(P).)

Mini test

15 minutes, without looking back at the lesson. Part I: circle one letter; Part II: write T or F for each statement; Part III: write the answer in the box.

Part I. Multiple choice

Question 1. In the system OxyzOxyz, a direction vector of the line d:{x=1−2ty=3z=−1+td: \begin{cases}x=1-2t\\ y=3\\ z=-1+t\end{cases} is
A. (1,3,−1)(1, 3, -1).B. (−2,3,1)(-2, 3, 1).C. (−2,0,1)(-2, 0, 1).D. (2,0,1)(2, 0, 1).
Question 2. Which point lies on the line x−12=y+2−1=z−33\dfrac{x-1}{2}=\dfrac{y+2}{-1}=\dfrac{z-3}{3}?
A. (−1,2,−3)(-1, 2, -3).B. (2,−1,3)(2, -1, 3).C. (3,−1,6)(3, -1, 6).D. (1,−2,3)(1, -2, 3).
Question 3. The line through A(1,0,2)A(1, 0, 2) perpendicular to (P):2x−y+z−3=0(P): 2x-y+z-3=0 has equations
A. x=1+2tx=1+2t, y=−ty=-t, z=2+tz=2+t.B. x=2+tx=2+t, y=−1y=-1, z=1+2tz=1+2t.C. x=1+2tx=1+2t, y=ty=t, z=2+tz=2+t.D. x=1+2tx=1+2t, y=−ty=-t, z=2−3tz=2-3t.
Question 4. Let d1:x=1+td_1: x=1+t, y=2ty=2t, z=3−tz=3-t and d2:x=sd_2: x=s, y=1+sy=1+s, z=2sz=2s. The lines d1d_1 and d2d_2 are
A. parallel.B. intersecting.C. skew.D. coincident.

Part II. True or false

Question 1. Let d:x=1+td: x=1+t, y=2−ty=2-t, z=−1+2tz=-1+2t and (P):x+y+z−4=0(P): x+y+z-4=0.
a) u⃗=(1,−1,2)\vec{u}=(1, -1, 2) is a direction vector of dd.T / F
b) dd is parallel to (P)(P).T / F
c) dd meets (P)(P) at the point (2,1,1)(2, 1, 1).T / F
d) dd is perpendicular to the plane (Q):x−y+2z+7=0(Q): x-y+2z+7=0. T / F

Part III. Short answer

Question 1. The line d:x=2+td: x=2+t, y=−1+2ty=-1+2t, z=3−tz=3-t meets the plane (P):x+y+z−10=0(P): x+y+z-10=0 at M(a,b,c)M(a, b, c). Find a+b+ca+b+c.
Question 2. Find the distance from M(2,3,1)M(2, 3, 1) to the line d:x=1+2td: x=1+2t, y=−1+ty=-1+t, z=1+2tz=1+2t (to 2 decimal places).

Score: ........ / 10

Hints (only when stuck)

Exercise 1.

What you need: x=x0+atx=x_0+at, y=y0+bty=y_0+bt, z=z0+ctz=z_0+ct.

Steps:

  1. Write the parametric equations.
  2. Substitute t=−1t=-1.
  3. Find tt from the x-coordinate of BB, then check the other two coordinates.
Exercise 2.

What you need: u⃗=AB→\vec{u}=\overrightarrow{AB}.

Steps:

  1. Find AB→\overrightarrow{AB}.
  2. Write both forms using the point AA.
Exercise 3.

What you need: if d⊥(P)d\perp(P) then u⃗d=n⃗P\vec{u}_d=\vec{n}_P; for the intersection, substitute the parametric equations into (P)(P).

Steps:

  1. Write dd with u⃗=(2,−1,3)\vec{u}=(2, -1, 3).
  2. Substitute into (P)(P) and solve for tt.
Exercise 4.

What you need: parallel lines share a direction vector; read u⃗\vec{u} from the denominators.

Steps:

  1. Read off u⃗=(2,−1,3)\vec{u}=(2, -1, 3).
  2. Write the equations through MM.
Exercise 5.

What you need: if the direction vectors are parallel, the lines are parallel or coincide; check one more point.

Steps:

  1. Read off u⃗1\vec{u}_1, u⃗2\vec{u}_2.
  2. Substitute MM into d1d_1 (find tt).
  3. Compare u⃗1\vec{u}_1 with n⃗P\vec{n}_P.
Exercise 6.

What you need: the position after tt seconds is (80t,60t,10t)(80t, 60t, 10t).

Steps:

  1. Solve 10t=150010t=1500.
  2. The point on the ground is (80t,60t,0)(80t, 60t, 0); find its distance from OO.
Exercise 7.

What you need: substitute the parametric equations into the equation of the plane.

Steps:

  1. Substitute into each plane and simplify in tt.
  2. Count the solutions.
Exercise 8.

What you need: if the direction vectors are not parallel, solve the system in tt, ss.

Steps:

  1. Compare (1,−1,2)(1, -1, 2) and (2,1,−1)(2, 1, -1).
  2. Solve the first two equations, then test the third.
Exercise 9.

What you need: u⃗=AB→\vec{u}=\overrightarrow{AB}; CC lies on ABAB when one value of tt works for all three coordinates.

Steps:

  1. AB→=(2,−1,2)\overrightarrow{AB}=(2, -1, 2).
  2. Solve 1+2t=71+2t=7 and check.
Exercise 10.

What you need: the projection HH of AA on (P)(P); HH is the midpoint of AA′AA'.

Steps:

  1. Find HH, then A′=2H−AA'=2H-A (in coordinates).
  2. Use the line A′BA'B and find where it meets (P)(P).

Self-check

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