1. Lesson 12 – 10 Vectors in Space
Pg. 664 #1–13 odd, 19, 22, 25, 27, 31, 32, 35, 37
Lesson 12 – 10 Vectors in Space
Pre-calculus
To apply vectors in 3-D space

2. Learning Objective
To apply vectors in 3-D space

3. Vectors in three dimensions can be considered from both a geometric and an algebraic approach.
Vectors in 3D
Most of the characteristics are extensions from vectors in the plane.
𝑣 = 𝑥 2 + 𝑦 2 + 𝑧 2
𝑣 = 𝑥, 𝑦, 𝑧
𝛼, 𝛽, 𝑎𝑛𝑑 𝛾 are the direction angles measured from the 𝑥, 𝑦, 𝑎𝑛𝑑 𝑧 axes.

4. Vectors in 3D The three unit vectors are 𝑖 = 1, 0, 0 𝑗 = 0, 1, 0
𝑘 = 0, 0, 1
𝑣 = 3, 4, 5
+ 5 𝑘
=3 𝑖
+ 4 𝑗

5. Vectors in 3D Vector Relationships
𝑣 = 𝑣 1 , 𝑣 2 , 𝑣 3 and 𝑤 = 𝑤 1 , 𝑤 2 , 𝑤 3 and 𝑎 is a scalar, then:
𝑣 + 𝑤 = 𝑣 1 + 𝑤 1 , 𝑣 2 + 𝑤 2 , 𝑣 3 + 𝑤 3
𝑣 − 𝑤 = 𝑣 1 − 𝑤 1 , 𝑣 2 − 𝑤 2 , 𝑣 3 − 𝑤 3
𝑎 𝑣 = 𝑎 𝑣 1 , 𝑎 𝑣 2 , 𝑎 𝑣 3
𝑣 ∙ 𝑤 = 𝑣 1 , 𝑣 2 , 𝑣 3 ∙ 𝑤 1 , 𝑤 2 , 𝑤 3
= 𝑣 1 𝑤 1 + 𝑣 2 𝑤 2 + 𝑣 3 𝑤 3
Unit vector in same direction as 𝑣 is:
𝑢 = 𝑣 1 𝑣 𝑖 + 𝑣 2 𝑣 𝑗 + 𝑣 3 𝑣 𝑘

6. Vectors in 3D 1. 𝑣 = 1, −2, 5 and 𝑤 = −2, 7, 11 and 𝑎=−2 and 𝑏=4
𝑤 − 𝑣 = −2, 7, 11 − 1, −2, 5
= −3, 9, 6
𝑣 ∙ 𝑤 = 1, −2, 5 ∙ −2, 7, 11
= 1 −2 + −2 7 +(5)(11)
=−2−14+55
=39
𝑎 𝑤 +𝑏 𝑣 =−2 −2, 7, , −2, 5
= 4, −14, −22 + 4, −8, 20
= 8, −22, −2

7. Vectors in 3D cos 𝜃 = 𝑣 ∙ 𝑤 𝑣 𝑤
More similarities with vectors in the plane:
Vectors in 3D
𝑣 ∙ 𝑤 = 𝑣 𝑤 cos 𝜃
cos 𝜃 = 𝑣 ∙ 𝑤 𝑣 𝑤
The vectors are perpendicular iff 𝑣 ∙ 𝑤 =0

8. Vectors in 3D cos 𝜃 = 𝑣 ∙ 𝑤 𝑣 𝑤
2. Determine the angle between vectors
𝑣 = 𝑖 − 5 𝑗 +2 𝑘 and 𝑤 =3 𝑖 + 𝑘
Vectors in 3D
cos 𝜃 = 1, − 5 , 2 ∙ 3, 0, (− 5 )
cos 𝜃 = 𝑣 ∙ 𝑤 𝑣 𝑤 =
= 5 10
cos 𝜃 = 1 2
θ= 𝑐𝑜𝑠 −
𝜃= 60 𝑜

9. Vectors in 3D 𝑄(𝑥, 𝑦, 𝑧) and 𝑃(𝑎, 𝑏, 𝑐)
Vector Equation of a Line (in space):
𝑄=𝑃+𝑡 𝑣
(𝑥, 𝑦, 𝑧)=(𝑎, 𝑏, 𝑐)+𝑡 𝑣 1 , 𝑣 2 , 𝑣 3
And therefore the parametric equations are:
𝑥=𝑎+𝑡 𝑣 1
𝑦=𝑏+𝑡 𝑣 2
𝑧=𝑐+𝑡 𝑣 3

10. 3. Determine both vector & parametric equations of the line in space determined by
𝑃(1, 4, 7) and 𝑅(−6, −4, −8)
Vectors in 3D
Direction vector:
𝑅−𝑃=
−6−1, −4−4, −8−7
= −7, −8, −15
Vector equation:
(𝑥, 𝑦, 𝑧)=
(1, 4, 7)+𝑡 −7,−8, −15
Parametric equations:
𝑥=1−7𝑡
𝑦=4−8𝑡
𝑧=7−15𝑡

11. Vectors in 3D Normal Vector:
- This is a vector perpendicular to a plane
- Meaning the vector is perpendicular to all the vectors that contains a specific point in the plane.
If 𝑃( 𝑥 0 , 𝑦 0 , 𝑧 0 ) and 𝑄 is (𝑥, 𝑦, 𝑧), and the normal is 𝑛 , then 𝑛 ∙ 𝑄−𝑃 =0

12. 4. Determine an equation of the plane containing the point 𝑃(1, 2, 3) with normal vector −2, 4, 7
Vectors in 3D
Let 𝑄(𝑥, 𝑦, 𝑧) represent all points in the plane
−2, 4, 7 ∙ 𝑥−1, 𝑦−2, 𝑧−3 =0
−2 𝑥− 𝑦− 𝑧−3 =0
−2𝑥+2+4𝑦−8+7𝑧−21=0
−2𝑥+4𝑦+7𝑧=27

13. Cross Product – An operation that applies to three–dimensional vectors, but NOT two–dimensional
Vectors in 3D
𝑣 𝑥 𝑢 = vector (not a scalar)
We will use determinants to help us solve.
Remember: 𝑎 𝑏 𝑐 𝑑 =𝑎𝑑−𝑏𝑐
𝑣 𝑥 𝑢 = 𝑖 𝑗 𝑘 𝑣 1 𝑣 2 𝑣 3 𝑢 1 𝑢 2 𝑢 3
Watch
= 𝑖 𝑗 𝑘 𝑣 1 𝑣 2 𝑣 3 𝑢 1 𝑢 2 𝑢 3
𝑖 𝑗 𝑘 𝑣 1 𝑣 2 𝑣 3 𝑢 1 𝑢 2 𝑢 3
𝑖 𝑗 𝑘 𝑣 1 𝑣 2 𝑣 3 𝑢 1 𝑢 2 𝑢 3
− 𝑣 1 𝑣 3 𝑢 1 𝑢 3 𝑗
+ 𝑣 1 𝑣 2 𝑢 1 𝑢 2 𝑘
= 𝑣 2 𝑣 3 𝑢 2 𝑢 3 𝑖

14. Vectors in 3D 𝑣 𝑥 𝑢 = 𝑖 𝑗 𝑘 𝑣 1 𝑣 2 𝑣 3 𝑢 1 𝑢 2 𝑢 3
𝑣 𝑥 𝑢 = 𝑖 𝑗 𝑘 𝑣 1 𝑣 2 𝑣 3 𝑢 1 𝑢 2 𝑢 3
Vectors in 3D
= 𝑣 2 𝑣 3 𝑢 2 𝑢 3 𝑖
− 𝑣 1 𝑣 3 𝑢 1 𝑢 3 𝑗
+ 𝑣 1 𝑣 2 𝑢 1 𝑢 2 𝑘
- This cross–product is perpendicular to the plane that the two vectors make.
- This fact enables us to find the equation of that plane.
(3 steps)
1) You will be given 3 points in a plane.
- Use them to create two vectors.
2) Find the normal to the plane using those two vectors.
3) Then create the equation of the plane using any of the three points.

15. 5. Find an equation of the plane containing the points 𝑃 2,−1, 6 , 𝑄(0,1, 5), and 𝑅(4, −1, 2)
Vectors in 3D
Step 1: Determine vectors 𝑃𝑄 and 𝑃𝑅
𝑄−𝑃= 0−2, 1−(−1), 5−6
= −2, 2, −1
𝑅−𝑃= 4−2, −1− −1 , 2−6
= 2, 0, −4

16. 5. Find an equation of the plane containing the points 𝑃 2,−1, 6 , 𝑄(0,1, 5), and 𝑅(4, −1, 2)
Vectors in 3D
Step 2: Cross Product: 𝑃𝑄 and 𝑃𝑅
𝑃𝑄 𝑥 𝑃𝑅 = 𝑖 𝑗 𝑘 −2 2 −1 2 0 −4
= 2 −1 0 −4 𝑖
− −2 −1 2 −4 𝑗
+ − 𝑘
=(−8−0) 𝑖
−(8− −2 ) 𝑗
+(0−4) 𝑘
=−8 𝑖
−10 𝑗
−4 𝑘
Normal to plane!

17. 5. Find an equation of the plane containing the points 𝑃 2,−1, 6 , 𝑄(0,1, 5), and 𝑅(4, −1, 2)
Vectors in 3D
Step 3: Equation of plane using 𝑃
−8,−10, −4 ∙ 𝑥−2, 𝑦+1, 𝑧−6 =0
−8 𝑥−2 −10 𝑦+1 −4 𝑧−6 =0
−8𝑥+16−10𝑦−10−4𝑧+24=0
𝑃 2,−1, 6
𝑄 𝑥,𝑦, 𝑧
−8𝑥−10𝑦−4𝑧=−30
Or 8𝑥+10𝑦+4𝑧=30

18. Assignment
Pg. 664 #1–13 odd, 19, 22, 25, 27, 31, 32, 35, 37