Last updated at March 11, 2017 by Teachoo

Transcript

Ex 11.3, 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (a) 2x + 3y + 4z – 12 = 0 Assume a point P(x1, y1, z1) on the given plane Since perpendicular to plane is parallel to normal vector Vector 𝑶𝑷 is parallel to normal vector 𝒏 to the plane. Given equation of plane is 2x + 3y + 4z − 12 = 0 2x + 3y + 4z = 12 Since, 𝑶𝑷 and 𝒏 are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional So, 𝑎1 𝑎2 = 𝑏1 𝑏2 = 𝑐1 𝑐2 = k 𝑥12 = 𝑦13 = 𝑧14 = k x1 = 2k , y1 = 3k , z1 = 4k Also, point P(x1, y1, z1) lies in the plane. Putting P (2k, 3k, 4k) in 2x + 3y + 4z = 12, 2(2k) + 3(3k) + 4(4k) = 12 4k + 9k + 16k = 12 29k = 12 ∴ k = 1229 So, 𝑥1 = 2k = 2 × 1229 = 2429 𝑦1 = 3k = 3 × 1229 = 3629 & 𝑧1 = 4k = 4 × 1229 = 4829 Therefore, coordinate of foot of perpendicular are 𝟐𝟒𝟐𝟗, 𝟑𝟔𝟐𝟗, 𝟒𝟖𝟐𝟗 Ex 11.3, 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (b) 3y + 4z – 6 = 0 Assume a point P(x1, y1, z1) on the given plane Since perpendicular to plane is parallel to normal vector Vector 𝑶𝑷 is parallel to normal vector 𝒏 to the plane. Given equation of plane is 3y + 4z − 6 = 0 3y + 4z = 6 0x + 3y + 4z = 6 Since, 𝑶𝑷 and 𝒏 are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional So, 𝑎1 𝑎2 = 𝑏1 𝑏2 = 𝑐1 𝑐2 = k 𝑥10 = 𝑦13 = 𝑧14 = k x1 = 0 , y1 = 3k , z1 = 4k Also, point P(x1, y1, z1) lies in the given plane. Putting P (0, 3k, 4k) in 0x + 3y + 4z = 6, 0 (k) + 3(3k) + 4(4k) = 6 25k = 6 k = 625 So, 𝑥1 = 0 𝑦1 = 3k = 3 × 6 25 = 1825 𝑧1 = 4k = 4 × 6 25 = 2425 Therefore, coordinates of foot of perpendicular are 𝟎, 𝟏𝟖𝟐𝟓, 𝟐𝟒𝟐𝟓 Ex 11.3, 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (c) x + y + z = 1 Assume a point P (x1, y1, z1) on the plane. Since perpendicular to plane is parallel to normal vector Vector 𝑶𝑷 is parallel to normal vector 𝒏 to the plane. Given, equation of the plane is x + y + z = 1 1x + 1y + 1z = 1 Since, 𝑶𝑷 and 𝒏 are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional So, 𝑎1 𝑎2 = 𝑏1 𝑏2 = 𝑐1 𝑐2 = k 𝑥11 = 𝑦11 = 𝑧11 = k x1 = y1 = z1 = k Also, point P(x1, y1, z1) lies in the given plane. Putting P (k, k, k) in x + y + z = 1, k + k + k = 1 3k = 1 ∴ k = 13 So, 𝑥1= k = 13 , 𝑦1 = k = 13 , 𝑧1= k = 13 Therefore, coordinate of foot of perpendicular are 𝟏𝟑, 𝟏𝟑, 𝟏𝟑 Ex 11.3, 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (d) 5y + 8 = 0 Assume a point P(x1, y1, z1) on the given plane Since perpendicular to plane is parallel to normal vector Vector 𝑶𝑷 is parallel to normal vector 𝒏 to the plane. Given, equation of plane 5y + 8 = 0 5y = − 8 − 5y = 8 0x − 5y + 0z = 8 Since, 𝑶𝑷 and 𝒏 are parallel their direction ratios are proportional. Finding direction ratios Since Direction ratios are proportional. 𝑎1 𝑎2 = 𝑏1 𝑏2 = 𝑐1 𝑐2 𝑥10 = 𝑦1 − 5 = 𝑧10 = k x1 = 0 , y1 = − 5k z1 = 0 Also, point P (x1, y1, z1) lies in the given plane. Putting (x1, y1, z1) in 0x − 5y + 0z = 8, 0x1 − 5y1 + 0z1 = 8, −5(−5k) = 8 25k = 8 ∴ k = 825 So, x1 = 0, y1 = –5k = − 5 × 825 = −85 z1 = 0 ∴ Coordinate of foot of perpendicular is 0, −𝟖𝟓, 0.

Example, 3
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Ex 11.1, 2 Important

Example, 6 Important

Example, 9 Important

Example 12 Important

Ex 11.2, 5 Important

Ex 11.2, 11 Important

Ex 11.2, 12 Important

Ex 11.2, 14 Important

Ex 11.2, 15 Important

Ex 11.2, 17 Important

Example 20 Important

Example 21 Important

Example 23 Important

Example 24 Important

Example, 25 Important

Ex 11.3, 4 Important You are here

Ex 11.3, 11 Important

Ex 11.3, 12 Important

Ex 11.3, 14 Important

Example 27 Important

Example 29 Important

Example 30 Important

Misc 6 Important

Misc 9 Important

Misc 14 Important

Misc 18 Important

Misc 20 Important

Misc 21 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.