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Grade 12 · Lesson 5.3

Angle formulas in space

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Goals

Remember

Exercises

Exercise 1

Find the angle between d1:x=1+td_1: x=1+t, y=2ty=2t, z=−tz=-t and d2:x−12=y−1=z+31d_2: \dfrac{x-1}{2}=\dfrac{y}{-1}=\dfrac{z+3}{1} (to one decimal place of a degree).

Exercise 2

Find the angle between the line d:x=1+td: x=1+t, y=ty=t, z=2z=2 and the plane (P):x+z−5=0(P): x+z-5=0.

Exercise 3

Find the angle between the planes (P):2x−y+z−1=0(P): 2x-y+z-1=0 and (Q):x+y+2z+3=0(Q): x+y+2z+3=0.

Exercise 4

Find the angle between the line Δ:x−12=y+1−2=z1\Delta: \dfrac{x-1}{2}=\dfrac{y+1}{-2}=\dfrac{z}{1} and the plane (Oxy)(Oxy) (to one decimal place of a degree).

Exercise 5

Let ABCD.A′B′C′D′ABCD.A'B'C'D' be a cube with edge 11, with A(0,0,0)A(0, 0, 0), B(1,0,0)B(1, 0, 0), D(0,1,0)D(0, 1, 0), A′(0,0,1)A'(0, 0, 1). Is each claim true or false?

a) The angle between the lines ACAC and B′D′B'D' is 90∘90^\circ.

b) The angle between the line A′CA'C and the plane (ABCD)(ABCD) is 45∘45^\circ.

c) The angle φ\varphi between the planes (A′BD)(A'BD) and (ABCD)(ABCD) satisfies cos⁡φ=13\cos\varphi=\dfrac{1}{\sqrt3}.

d) The angle between the lines AC′AC' and BDBD is 60∘60^\circ.

Exercise 6

A wheelchair ramp is an inclined plane: for every 1212 m along the ground it rises 11 m. Take the ramp as the plane (P):x−12z=0(P): x-12z=0 (unit: metre). Find the angle between the ramp and the ground (in degrees, to 1 decimal place).

Exercise 7

The pyramid S.ABCDS.ABCD has a rectangular base with AB=3AB=3, AD=4AD=4, SASA perpendicular to the base and SA=5SA=5. Find the angle between SCSC and the base, and the angle between SCSC and the plane (SAB)(SAB) (to one decimal place of a degree).

Exercise 8

(Revisiting The equation of a plane.) What angle does the plane through A(2,0,0)A(2, 0, 0), B(0,1,0)B(0, 1, 0), C(0,0,2)C(0, 0, 2) make with the plane (Oxy)(Oxy) (to one decimal place of a degree)?

Exercise 9

(Revisiting The equation of a line in space.) Let A(1,2,−1)A(1, 2, -1), B(3,−1,1)B(3, -1, 1) and (P):x−2y+2z+3=0(P): x-2y+2z+3=0. Find the angle between the line ABAB and (P)(P) (to one decimal place of a degree).

Exercise 10 Challenge

Let ABCD.A′B′C′D′ABCD.A'B'C'D' be a cube with edge 11. Find the angle between the diagonal BD′BD' and the plane (A′BD)(A'BD) (to one decimal place of a degree).

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. The angle between two lines with direction vectors u⃗1=(1,0,1)\vec{u}_1=(1, 0, 1) and u⃗2=(0,−1,−1)\vec{u}_2=(0, -1, -1) is
A. 60∘60^\circ.B. 120∘120^\circ.C. 30∘30^\circ.D. 45∘45^\circ.
Question 2. The angle between a line with direction vector u⃗=(1,1,0)\vec{u}=(1, 1, 0) and a plane with normal vector n⃗=(0,1,1)\vec{n}=(0, 1, 1) is
A. 60∘60^\circ.B. 45∘45^\circ.C. 30∘30^\circ.D. 90∘90^\circ.
Question 3. Let φ\varphi be the angle between the plane (P):x−2y+2z=0(P): x-2y+2z=0 and the plane (Oxy)(Oxy). Then
A. sin⁡φ=23\sin\varphi=\dfrac23.B. cos⁡φ=23\cos\varphi=\dfrac23.C. cos⁡φ=29\cos\varphi=\dfrac29.D. cos⁡φ=−23\cos\varphi=-\dfrac23.
Question 4. In the cube ABCD.A′B′C′D′ABCD.A'B'C'D', the angle between the lines A′C′A'C' and B′CB'C is
A. 45∘45^\circ.B. 90∘90^\circ.C. 60∘60^\circ.D. 120∘120^\circ.

Part II. True or false

Question 1. Let (P):2x+y−2z+1=0(P): 2x+y-2z+1=0 and the line d:x=1+td: x=1+t, y=−2ty=-2t, z=3z=3.
a) The angle between dd and (P)(P) is 0∘0^\circ.T / F
b) (P)(P) is perpendicular to the plane (Q):x+z−4=0(Q): x+z-4=0.T / F
c) The angle between (P)(P) and (Oxy)(Oxy) is about 41.8∘41.8^\circ.T / F
d) dd is perpendicular to the plane (Oxy)(Oxy). T / F

Part III. Short answer

Question 1. Find the angle between the planes (P):x+2y+2z−1=0(P): x+2y+2z-1=0 and (Q):2x+y−z+5=0(Q): 2x+y-z+5=0, in degrees, to 1 decimal place.
Question 2. A filming drone at D(0,0,40)D(0, 0, 40) points its camera at a target T(30,40,0)T(30, 40, 0) on the ground (unit: metre). Find the angle between the line of sight DTDT and the ground, in degrees, to 1 decimal place.

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

Hints (only when stuck)

Exercise 1.

What you need: cos⁡φ=∣u⃗1⋅u⃗2∣∣u⃗1∣∣u⃗2∣\cos\varphi=\dfrac{|\vec{u}_1\cdot\vec{u}_2|}{|\vec{u}_1||\vec{u}_2|}.

Steps:

  1. Read off u⃗1=(1,2,−1)\vec{u}_1=(1, 2, -1), u⃗2=(2,−1,1)\vec{u}_2=(2, -1, 1).
  2. Compute the scalar product and the two lengths.
Exercise 2.

What you need: sin⁡φ=∣u⃗⋅n⃗∣∣u⃗∣∣n⃗∣\sin\varphi=\dfrac{|\vec{u}\cdot\vec{n}|}{|\vec{u}||\vec{n}|}.

Steps:

  1. u⃗=(1,1,0)\vec{u}=(1, 1, 0), n⃗=(1,0,1)\vec{n}=(1, 0, 1).
  2. Compute sin⁡φ\sin\varphi.
Exercise 3.

What you need: cos⁡φ=∣n⃗1⋅n⃗2∣∣n⃗1∣∣n⃗2∣\cos\varphi=\dfrac{|\vec{n}_1\cdot\vec{n}_2|}{|\vec{n}_1||\vec{n}_2|}.

Steps:

  1. n⃗1=(2,−1,1)\vec{n}_1=(2, -1, 1), n⃗2=(1,1,2)\vec{n}_2=(1, 1, 2).
  2. Compute the cosine.
Exercise 4.

What you need: (Oxy)(Oxy) has normal vector k⃗=(0,0,1)\vec{k}=(0, 0, 1).

Steps:

  1. u⃗=(2,−2,1)\vec{u}=(2, -2, 1), ∣u⃗∣=3|\vec{u}|=3.
  2. Compute sin⁡φ\sin\varphi.
Exercise 5.

What you need: the three angle formulas; (A′BD):x+y+z−1=0(A'BD): x+y+z-1=0 (intercept form).

Steps:

  1. Write the vectors AC→\overrightarrow{AC}, B′D′→\overrightarrow{B'D'}, A′C→\overrightarrow{A'C}, AC′→\overrightarrow{AC'}, BD→\overrightarrow{BD} in coordinates.
  2. Use the right formula for each kind of angle.
Exercise 6.

What you need: the angle between two planes; the ground is (Oxy)(Oxy).

Steps:

  1. n⃗P=(1,0,−12)\vec{n}_P=(1, 0, -12), k⃗=(0,0,1)\vec{k}=(0, 0, 1).
  2. Compute the cosine, then the angle.
Exercise 7.

What you need: put AA at the origin; (ABCD)(ABCD) is (Oxy)(Oxy) and (SAB)(SAB) is (Ozx):y=0(Ozx): y=0.

Steps:

  1. B(3,0,0)B(3, 0, 0), D(0,4,0)D(0, 4, 0), S(0,0,5)S(0, 0, 5), C(3,4,0)C(3, 4, 0).
  2. SC→=(3,4,−5)\overrightarrow{SC}=(3, 4, -5); use sin⁡φ\sin\varphi with (0,0,1)(0, 0, 1) and (0,1,0)(0, 1, 0).
Exercise 8.

What you need: the intercept form; the angle between two planes.

Steps:

  1. (ABC):x2+y+z2=1  ⟺  x+2y+z−2=0(ABC): \dfrac{x}{2}+y+\dfrac{z}{2}=1\iff x+2y+z-2=0.
  2. Use n⃗=(1,2,1)\vec{n}=(1, 2, 1) and k⃗=(0,0,1)\vec{k}=(0, 0, 1).
Exercise 9.

What you need: the line ABAB has direction vector AB→\overrightarrow{AB}; sin⁡φ=∣u⃗⋅n⃗∣∣u⃗∣∣n⃗∣\sin\varphi=\dfrac{|\vec{u}\cdot\vec{n}|}{|\vec{u}||\vec{n}|}.

Steps:

  1. AB→=(2,−3,2)\overrightarrow{AB}=(2, -3, 2).
  2. Compute sin⁡φ\sin\varphi with n⃗=(1,−2,2)\vec{n}=(1, -2, 2).
Exercise 10.

What you need: use the axes of exercise 5; (A′BD):x+y+z−1=0(A'BD): x+y+z-1=0.

Steps:

  1. B(1,0,0)B(1, 0, 0), D′(0,1,1)D'(0, 1, 1), so BD′→=(−1,1,1)\overrightarrow{BD'}=(-1, 1, 1).
  2. Compute sin⁡φ\sin\varphi with n⃗=(1,1,1)\vec{n}=(1, 1, 1).

Self-check

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