How do you calculate angle between 2 lines in 3 dimensional geometry? Learn this and much more in this presentation/video. Here, we learn how to calculate angle between 2 lines, equation of a line passing through 2 points and perpendicular to a line and how to find the foot of a perpendicular from a point to a line. These are useful for grade 12 mathematics students and students appearing for the NATA eligibility test in architecture which has maths as a section. For more videos, subscribe to my channel or visit my page https://www.mathmadeeasy.co/about

1. UCjmCXXIjd03JQad8-rUzs0Q
GRADE 12 MATHS
NATA eligibility test (mathematics)

2. Angle between 2 lines
3 dimensional geometry

3.  How do you calculate the angle between 2 lines?
 Why do we need it?
 This is a part of 3 dimensional geometry and used in grade 12 mathematics
 Also used in the maths section of the NATA eligibility test in architecturee

4. We learn to calculate
Angle between 2 lines
Equation of a line passing through a point and perpendicular to 2 lines
How to calculate the foot of a perpendicular from a point to a line

5. (𝑥1, 𝑦1, 𝑧1)
𝑛

6. Given 2 lines in Cartesian form
𝑥 − 𝑥1
𝑎1
=
𝑦 − 𝑦1
𝑏1
=
𝑧 − 𝑧1
𝑐1
and
𝑥 − 𝑥2
𝑎2
=
𝑦 − 𝑦2
𝑏2
=
𝑧 − 𝑧2
𝑐2
Angle between the 2 lines 𝑐𝑜𝑠𝜃 =
𝑎1 𝑎2+𝑏1 𝑏2+𝑐1 𝑐2
𝑎1
2+𝑏1
2+𝑐1
2 𝑎2
2+𝑏2
2+𝑐2
2
Acute angle
𝑐𝑜𝑠𝜃 =
𝑎1 𝑎2 + 𝑏1 𝑏2 + 𝑐1 𝑐2
𝑎1
2
+ 𝑏1
2
+ 𝑐1
1
𝑎2
2
+ 𝑏2
2
+ 𝑐2
2

7. 𝐼𝑓 < 𝑙1, 𝑚1, 𝑛1 >, < 𝑙2, 𝑚2, 𝑛2 > 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑠𝑖𝑛𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒
𝑐𝑜𝑠𝜃 = 𝑙1 𝑙2 + 𝑚1 𝑚2 + 𝑛1 𝑛2

8. Q 1 Find the angle between the lines 𝑟=(3 𝑖 − 2 𝑗 + 6 𝑘) + 𝛿(2 𝑖 + 𝑗 + 2 𝑘)
𝑎𝑛𝑑 𝑟 = 2 𝑗 − 5 𝑘 + 𝜎(6 𝑖 + 3 𝑗 + 2 𝑘)

9. Converting to Cartesian form
𝑥 − 3
2
=
𝑦 + 2
1
=
𝑧 − 6
2
𝑥
6
=
𝑦 − 2
3
=
𝑧 + 5
2
𝑐𝑜𝑠𝜃 =
12 + 3 + 4
4 + 1 + 4 36 + 9 + 4
=
19
21
𝜃=𝑐𝑜𝑠−1
(
19
21
)

10. Equation of a line passing through a point 𝑎 𝑎𝑛𝑑 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜
the lines 𝑟1 = 𝑎1 + 𝛼𝑏1 𝑎𝑛𝑑 𝑟2 = 𝑎2 + 𝜇𝑏2
𝑟=𝑎 +𝜌(𝑏1 × 𝑏2)

12. Q 2) Find the equation of the line passing through (1,2,-4) and
perpendicular to the lines
𝑟=(8 𝑖 − 19 𝑗 + 10 𝑘) + 𝛼(3 𝑖 − 16 𝑗 + 7 𝑘) 𝑎𝑛𝑑
𝑟 = 15 𝑖 + 29 𝑗 + 5 𝑘 + 𝜇(3 𝑖 + 8 𝑗 − 5 𝑘)

13. The required line is perpendicular to the 2 lines
𝑖 𝑗 𝑘
3 −16 7
3 8 −5
= 𝑖 80 − 56 − 𝑗 −15 − 21 + 𝑘(24 + 48)
= 24 𝑖 + 36 𝑗 + 72 𝑘

14. Equation of the required line 𝑖+2 𝑗 − 4 𝑘 + 𝑡(24 𝑖 + 36 𝑗 + 72 𝑘)
= 1 + 24𝑡 𝑖 + 2 + 36𝑡 𝑗 + (−4 + 72𝑡) 𝑘

15. 𝑥 + 3
5
=
𝑦 − 1
2
=
𝑧 + 4
3
Q 3 Find the foot of the perpendicular drawn from (0,2,3) on the line

16. 𝑙𝑒𝑡
𝑥 + 3
5
=
𝑦 − 1
2
=
𝑧 + 4
3
= 𝑡
x = 5t -3 , y = 2t +1, z = 3t -4
(0,2,3)
(5t -3, 2t+1,3t-4)
Direction ratios of the perpendicular
< 5t-3, 2t-1, 3t-7>
Direction ratios of the line =<5,2,3>

17. Since the 2 lines are perpendicular
5(5t-3)+2(2t-1)+3(3t-7)=0
t = 1
Point of intersection (2,3, -1)

18. What have we learnt?
How to calculate angle between 2 lines
How to calculate equation of a line passing through a point and
perpendicular to 2 lines
How to calculate foot of the perpendicular from a point to a line

19. Was this useful?
Subscribe to my channel
https://www.mathmadeeasy.co/about
View my playlist
UCjmCXXIjd03JQad8-rUzs0Q
https://www.mathmadeeasy.co/about-3
https://www.youtube.com/playlist?list=PL26R7TjUyi8zn1C9-
TWlECndvq6hIs5gh