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Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

Ex 11.3, 4 - Chapter 11 Class 12 Three Dimensional Geometry - Part 11

Ex 11.3, 4 - Chapter 11 Class 12 Three Dimensional Geometry - Part 12
Ex 11.3, 4 - Chapter 11 Class 12 Three Dimensional Geometry - Part 13 Ex 11.3, 4 - Chapter 11 Class 12 Three Dimensional Geometry - Part 14

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 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 is parallel to normal vector to the plane. Given, equation of plane 5y + 8 = 0 5y = 8 5y = 8 0x 5y + 0z = 8 Since, and are parallel their direction ratios are proportional. Finding direction ratios Since Direction ratios are proportional. 1 2 = 1 2 = 1 2 1 0 = 1 5 = 1 0 = 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 = 8 25 So, x1 = 0, y1 = 5k = 5 8 25 = 8 5 z1 = 0 Coordinate of foot of perpendicular is 0, , 0 .

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.