![From (ii) and (iii) , by cross multiplication, we get
𝑎
1 − 4
=
𝑏
8 − 1
=
𝑐
1 − 2
𝑜𝑟,
𝑎
−3
=
𝑏
7
=
𝑐
−1
= 𝑘 (𝑠𝑎𝑦)
Therefore, 𝑎 = −3𝑘, 𝑏 = 7𝑘 𝑎𝑛𝑑 𝑐 = −𝑘, 𝑤ℎ𝑒𝑟𝑒 𝑘 ≠ 0
Putting the values of a, b and c in (i), we get
−3𝑘 𝑥 − 1 + 7𝑘 𝑦 − 𝑘 𝑧 + 1 = 0
𝑜𝑟, 𝑘[−3 𝑥 − 1 + 7 𝑦 − 𝑧 + 1 ] = 0
𝑜𝑟, −3 𝑥 − 1 + 7 𝑦 − 𝑧 + 1 = 0
𝑜𝑟, −3𝑥 + 3 + 7𝑦 − 𝑧 − 1 = 0
𝑜𝑟, −3𝑥 + 7𝑦 − 𝑧 + 2 = 0
𝑜𝑟, 3𝑥 − 7𝑦 + 𝑧 − 2 = 0](https://image.slidesharecdn.com/plane-200904121212/85/Plane-in-3-dimensional-geometry-22-320.jpg)
![From (ii) and (iii) , by cross multiplication, we get
𝑎
−12 + 4
=
𝑏
−12 − 12
=
𝑐
2 + 6
𝑜𝑟,
𝑎
−8
=
𝑏
−24
=
𝑐
8
𝑜𝑟,
𝑎
1
=
𝑏
3
=
𝑐
−1
= 𝑘 (𝑠𝑎𝑦)
Therefore, 𝑎 = 𝑘, 𝑏 = 3𝑘 𝑎𝑛𝑑 𝑐 = −𝑘, 𝑤ℎ𝑒𝑟𝑒 𝑘 ≠ 0
Putting the values of a, b and c in (i), we get
𝑘 𝑥 − 1 + 3𝑘 (𝑦 − 1) − 𝑘 𝑧 − 2 = 0
𝑜𝑟, 𝑘[ 𝑥 − 1 + 3 (𝑦 − 1) − 𝑧 − 2 ] = 0
𝑜𝑟, 𝑥 − 1 + 3 (𝑦 − 1) − 𝑧 − 2 = 0
𝑜𝑟, 𝑥 − 1 + 3𝑦 − 3 − 𝑧 + 2 = 0
𝑜𝑟, 𝑥 + 3𝑦 − 𝑧 − 2 = 0](https://image.slidesharecdn.com/plane-200904121212/85/Plane-in-3-dimensional-geometry-24-320.jpg)
Plane in 3 dimensional geometry
- 2. Definition A plane is a flat, two dimensional surface that extends infinitely far.
- 5. Formula 1) General equation of a plane that passes through a given point (𝑥1, 𝑦1, 𝑧1) is 𝑎 𝑥 − 𝑥1 + 𝑏 𝑦 − 𝑦1 + 𝑐 𝑧 − 𝑧1 = 0 2) The equation of a plane passing through three non-collinear points 𝑥1, 𝑦1, 𝑧1 , 𝑥2, 𝑦2, 𝑧2 and 𝑥3, 𝑦3, 𝑧3 𝑥 𝑦 𝑧 1 𝑥1 𝑦1 𝑧1 1 𝑥2 𝑦2 𝑧2 1 𝑥3 𝑦3 𝑧3 1 = 0
- 6. Formula 3) The equation of a plane parallel to 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑 = 0 is 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑘 = 0 4) The distance between the parallel planes 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 and 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑1 = 0 is 𝑑 − 𝑑1 𝑎2 + 𝑏2 + 𝑐2 5) The distance of the plane 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑 = 0 from the point (𝑥1, 𝑦1, 𝑧1) is 𝑎𝑥1 + 𝑏𝑦1 + 𝑐𝑧1 + 𝑑 𝑎2 + 𝑏2 + 𝑐2
- 7. Formula 6) The equation of a plane in the intercept form is 𝑥 𝑎 + 𝑦 𝑏 + 𝑧 𝑐 = 1 7) The equation of the plane passing through the line of intersection of two given planes 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 … … (1) 𝑎𝑛𝑑 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑦 + 𝑑1 = 0 … … 2 is given by the following equation 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 + 𝑘 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑦 + 𝑑1 = 0
- 8. Formula 8. The angle between two planes 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 and 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑧 + 𝑑1 = 0 is cos 𝜃 = 𝑎𝑎1 + 𝑏𝑏1 + 𝑐𝑐1 𝑎2+𝑏2 + 𝑐2 𝑎1 2 + 𝑏1 2 + 𝑐1 2 Condition of perpendicular: 𝑎𝑎1 + 𝑏𝑏1 + 𝑐𝑐1 = 0 Condition of parallel: a a1 = b b1 = c c1
- 9. 1. Find the equation of the plane with intercepts 2,3 and 4 on the x, y and z axis respectively
- 10. 2. Find the distance between the parallel planes (a) 2𝑥 − 2𝑦 + 𝑧 + 3 = 0 and 4𝑥 − 4𝑦 + 2𝑧 + 5 = 0 (b) 4𝑥 − 3𝑦 − 12𝑧 + 6 = 0 and 4𝑥 − 3𝑦 − 12𝑧 − 4 = 0 c) 𝑥 − 𝑦 − 2𝑧 − 3 = 0 and 2𝑥 − 2𝑦 − 4𝑧 − 6 = 0 Solution: a)Given that 2𝑥 − 2𝑦 + 𝑧 + 3 = 0 … . (𝑖) and 4𝑥 − 4𝑦 + 2𝑧 + 5 = 0 or, 2𝑥 − 2𝑦 + 𝑧 + 5 2 = 0 … … (𝑖𝑖) Here 𝑎 = 2, 𝑏 = −2, 𝑐 = 1, 𝑑 = 3 𝑎𝑛𝑑 𝑑1 = 5 2 Then the required distance = 𝑑 − 𝑑1 𝑎2 + 𝑏2 + 𝑐2 = 3 − 5 2 4 + 4 + 1 = 6 − 5 2 9 = 1 2 3 = 1 6
- 11. 3. a) Prove that the points 1, −1,3 and (3,3,3) are equidistant from the plane 5𝑥 + 2𝑦 − 7𝑧 + 9 = 0 and on opposite side of it. b) Find the distance of the points 2, 0, 1 and (3, −3, 2) from the plane 𝑥 − 2𝑦 + 𝑧 = 6 and find whether the two points lie on the same side or opposite sides of the plane. Solution: Given plane is 5𝑥 + 2𝑦 − 7𝑧 + 9 = 0 … … (𝑖) The distance of (i) from 1, −1,3 is 5 × 1 + 2 × −1 − 7 × 3 + 9 52 + 22 + (−7)2 = −9 78 = 9 78 The distance of (i) from 3,3,3 is 5 × 3 + 2 × 3 − 7 × 3 + 9 52 + 22 + (−7)2 = 9 78 = 9 78
- 12. Therefore, the points 1, −1,3 and (3,3,3) are equidistant from the plane 5𝑥 + 2𝑦 − 7𝑧 + 9 = 0 (proved) Putting the point 1, −1,3 on the L.H.S of (i), we get 5 × 1 + 2 × −1 − 7 × 3 + 9 = −9, negative value Putting the point 3,3,3 on the L.H.S of (i), we get 5 × 3 + 2 × 3 − 7 × 3 + 9 = 9, positive value The two results are of opposite signs. Therefore, the given two points lie on the opposite sides of the plane (proved)
- 13. 4. Find the angles between the two planes a) 3𝑥 − 6𝑦 + 2𝑧 = 7 and 2𝑥 + 2𝑦 − 2𝑧 = 5 , b) 2x-y+z=6 and x+y+2z=7 Solution: a) Here 𝑎 = 3, 𝑏 = −6, 𝑐 = 2 𝑎𝑛𝑑𝑎1 = 2, 𝑏1 = 2, 𝑐1 = −2 COS 𝜃 = 3 × 2 + −6 × 2 + 2 × (−2) 32+(−6)2+22 22 + 22 + (−2)2 = −10 7 × 2 3 = 5 7 3 Therefore, 𝜃 = COS−1 5 7 3
- 14. 5. Find the equation of the plane through the points (a) (2,1,3), (-1,-2,4), (4,2,1) , (b) (1,1,-1), (-2,-2,2), (-1,-1,2) (c) (2,1,-3),(3,-1,4), (7,5,6), (d) (2,3,1), (1,1,3), (2,2,3) Solution: (a) The required equation of plane is 𝑥 𝑦 𝑧 1 𝑥1 𝑦1 𝑧1 1 𝑥2 𝑦2 𝑧2 1 𝑥3 𝑦3 𝑧3 1 = 0, 𝑜𝑟, 𝑥 𝑦 𝑧 1 2 1 3 1 −1 − 2 4 1 4 2 1 1 = 0 𝑜𝑟, 𝑥 − 2 𝑦 − 1 𝑧 − 1 1 − 1 2 + 1 1 + 2 3 − 4 1 − 1 −1 − 4 − 2 − 2 4 − 1 1 − 1 4 2 1 1 = 0 𝑟1 ′ → 𝑟1 − 𝑟2 𝑟2 ′ → 𝑟2 − 𝑟3 𝑟3 ′ → 𝑟3 − 𝑟4
- 15. 𝑜𝑟, 𝑥 − 2 𝑦 − 1 𝑧 − 1 0 3 3 − 1 0 −5 − 4 3 0 4 2 1 1 = 0 𝑜𝑟, 1. 𝑥 − 2 𝑦 − 1 𝑧 − 1 3 3 −1 −5 −4 3 = 0 𝑜𝑟, 𝑥 − 2 9 − 4 − 𝑦 − 1 9 − 5 + 𝑧 − 1 −12 + 15 = 0 𝑜𝑟, 𝑥 − 2 5 − 𝑦 − 1 4 + 𝑧 − 1 3 = 0 𝑜𝑟, 5𝑥 − 10 − 4𝑦 + 4 + 3𝑧 − 3 = 0 𝑜𝑟, 5𝑥 − 4𝑦 + 3𝑧 − 9 = 0
- 16. 6. Find the equation of the plane passing through the intersection of the planes a) 𝑥 − 2𝑦 + 3𝑧 + 4 = 0 and 2𝑥 − 3𝑦 + 4𝑧 − 7 = 0 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 1, −1, 1 . b) 𝑥 + 2𝑦 + 3𝑧 + 4 = 0 and 4𝑥 + 3𝑦 + 2𝑧 + 1 = 0 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 1, 2, 3 . Solution: a) The equation of the plane through the intersection of the given planes is 𝑥 − 2𝑦 + 3𝑧 + 4 + 𝑘 2𝑥 − 3𝑦 + 4𝑧 − 7 = 0 … … (𝑖) Since the plane (i) passes through the point (1, −1, 1) , we get 1+2 + 3 + 4 + 𝑘 2 + 3 + 4 − 7 = 0 𝑜𝑟, 10 + 2𝑘 = 0 𝑜𝑟, 𝑘 = −5
- 17. Putting the value of k in (i), we get 𝑥 − 2𝑦 + 3𝑧 + 4 − 5 2𝑥 − 3𝑦 + 4𝑧 − 7 = 0 𝑜𝑟, 𝑥 − 2𝑦 + 3𝑧 + 4 − 10𝑥 + 15𝑦 − 20𝑧 + 35 = 0 𝑜𝑟, −9𝑥 + 13𝑦 − 17𝑧 + 39 = 0 𝑜𝑟, 9𝑥 − 13𝑦 + 17𝑧 − 39 = 0
- 18. 7. Find the equation of the plane through a) the point (4,0,1) and parallel to the plane 4𝑥 + 3𝑦 − 11𝑧 + 6 = 0 b) the point (1,2,2) and parallel to the plane 3𝑥 + 2𝑦 + 𝑧 = 7 Solution: a) The equation of the plane parallel to 4𝑥 + 3𝑦 − 11𝑧 + 6 = 0 is given by 4𝑥 + 3𝑦 − 11𝑧 + 𝑘 = 0 … … (𝑖) Since the plane (i) passes through the point (4,0,1) , we get 16 + 0 − 11 + 𝑘 = 0 𝑜𝑟, 𝑘 + 5 = 0 𝑜𝑟, 𝑘 = −5 Putting the value of k in (i), we get 4𝑥 + 3𝑦 − 11𝑧 − 5 = 0
- 19. 8. Find the equation of the plane which is parallel to the plane a) 4𝑥 − 4𝑦 + 7𝑧 − 3 = 0 and distance 4 units from the point (3,1, −2) b) 2𝑥 − 3𝑦 − 6𝑧 − 14 = 0 and distance 5 units from the origin. Solution: Given plane is 4𝑥 − 4𝑦 + 7𝑧 − 3 = 0 … … (𝑖) The equation of plane parallel to the plane (i) is given by 4𝑥 − 4𝑦 + 7𝑧 + 𝑘 = 0 … … (𝑖𝑖) The distance of the plane (ii) from the point (3,1, −2) is 4 × 3 − 4 × 1 + 7 × (−2) + 𝑘 16 + 16 + 49 = 12 − 4 − 14 + 𝑘 81 = 𝑘 − 6 9
- 20. According to question 𝑘 − 6 9 = 4 or, 𝑘−6 9 = ±4 or, 𝑘 − 6 = ±36 or, 𝑘 = 6 ± 36 = 42, −30 Putting the values of k in (ii), we get 4𝑥 − 4𝑦 + 7𝑧 + 42 = 0 and 4 𝑥 − 4𝑦 + 7𝑧 − 32 = 0
- 21. 9) Find the equation of the plane which passes through a) The points 1,0, −1 𝑎𝑛𝑑 (2,1,3) and is perpendicular to the plane 2𝑥 + 𝑦 + 𝑧 = 1 b) The points 2,2,1 𝑎𝑛𝑑 (9,3,6) and is perpendicular to the plane 2𝑥 + 6𝑦 + 6𝑧 = 9 Solution: a) The equation of plane through the point 1,0, −1 is 𝑎 𝑥 − 1 + 𝑏 𝑦 − 0 + 𝑐 𝑧 + 1 = 0 or, 𝑎 𝑥 − 1 + 𝑏𝑦 + 𝑐 𝑧 + 1 = 0 … … (𝑖) Since the plane (i) passes through (2,1,3) , then it becomes 𝑎 + 𝑏 + 4𝑐 = 0 … … (𝑖𝑖) Also the plane (i) is perpendicular to the plane 2𝑥 + 𝑦 + 𝑧 = 1 Therefore, 2𝑎 + 𝑏 + 𝑐 = 0 … … (𝑖𝑖𝑖)
- 22. From (ii) and (iii) , by cross multiplication, we get 𝑎 1 − 4 = 𝑏 8 − 1 = 𝑐 1 − 2 𝑜𝑟, 𝑎 −3 = 𝑏 7 = 𝑐 −1 = 𝑘 (𝑠𝑎𝑦) Therefore, 𝑎 = −3𝑘, 𝑏 = 7𝑘 𝑎𝑛𝑑 𝑐 = −𝑘, 𝑤ℎ𝑒𝑟𝑒 𝑘 ≠ 0 Putting the values of a, b and c in (i), we get −3𝑘 𝑥 − 1 + 7𝑘 𝑦 − 𝑘 𝑧 + 1 = 0 𝑜𝑟, 𝑘[−3 𝑥 − 1 + 7 𝑦 − 𝑧 + 1 ] = 0 𝑜𝑟, −3 𝑥 − 1 + 7 𝑦 − 𝑧 + 1 = 0 𝑜𝑟, −3𝑥 + 3 + 7𝑦 − 𝑧 − 1 = 0 𝑜𝑟, −3𝑥 + 7𝑦 − 𝑧 + 2 = 0 𝑜𝑟, 3𝑥 − 7𝑦 + 𝑧 − 2 = 0
- 23. 10. Find the equation of the planes a) through (1,1,2) and perpendicular to each of the planes 2𝑥 − 2𝑦 − 4𝑧 − 6 = 0 𝑎𝑛𝑑 3𝑥 + 𝑦 + 6𝑧 − 4 = 0 b) through (−1,3,2) and perpendicular to each of the planes 𝑥 + 3𝑦 + 2𝑧 = 5 𝑎𝑛𝑑 3𝑥 + 3𝑦 + 2𝑧 = 8 Solution: a) The equation of plane through the point 1,1,2 is 𝑎 𝑥 − 1 + 𝑏 𝑦 − 1 + 𝑐 𝑧 − 2 = 0 … … (𝑖) Since the plane (i) is perpendicular to each of the planes 2𝑥 − 2𝑦 − 4𝑧 − 6 = 0 𝑎𝑛𝑑 3𝑥 + 𝑦 + 6𝑧 − 4 = 0 We get 2𝑎 − 2𝑏 − 4𝑐 = 0 … … (𝑖𝑖) and 3𝑎 + 𝑏 + 6𝑐 = 0 … … (𝑖𝑖𝑖)
- 24. From (ii) and (iii) , by cross multiplication, we get 𝑎 −12 + 4 = 𝑏 −12 − 12 = 𝑐 2 + 6 𝑜𝑟, 𝑎 −8 = 𝑏 −24 = 𝑐 8 𝑜𝑟, 𝑎 1 = 𝑏 3 = 𝑐 −1 = 𝑘 (𝑠𝑎𝑦) Therefore, 𝑎 = 𝑘, 𝑏 = 3𝑘 𝑎𝑛𝑑 𝑐 = −𝑘, 𝑤ℎ𝑒𝑟𝑒 𝑘 ≠ 0 Putting the values of a, b and c in (i), we get 𝑘 𝑥 − 1 + 3𝑘 (𝑦 − 1) − 𝑘 𝑧 − 2 = 0 𝑜𝑟, 𝑘[ 𝑥 − 1 + 3 (𝑦 − 1) − 𝑧 − 2 ] = 0 𝑜𝑟, 𝑥 − 1 + 3 (𝑦 − 1) − 𝑧 − 2 = 0 𝑜𝑟, 𝑥 − 1 + 3𝑦 − 3 − 𝑧 + 2 = 0 𝑜𝑟, 𝑥 + 3𝑦 − 𝑧 − 2 = 0