Example 16 - Chapter 11 Class 12 Three Dimensional Geometry
Last updated at Dec. 24, 2019 by Teachoo
Transcript
Example 16 Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 4z – 6 = 0. Let point P(x1, y1, z1) be foot of perpendicular from origin Since perpendicular to plane is parallel to normal vector Vector 𝑶𝑷 is parallel to normal vector 𝒏 Given equation of the plane is 2x − 3y + 4z − 6 = 0 2x − 3y + 4z = 6 Since, 𝑶𝑷 and 𝒏 are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional 𝑎1 𝑎2 = 𝑏1 𝑏2 = 𝑐1 𝑐2 = k 𝑥12 = 𝑦1 − 3 = 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 = 6 2(2k) − 3(−3k) + 4(4k) = 12 4k + 9k + 16k = 6 29k = 6 ∴ k = 629 So, 𝑥1 = 2k = 2 × 629 = 1229 𝑦1 = –3k = −3 × 6 29 = −1829 & 𝑧1 = 4k = 4 × 629 = 2429 Therefore, coordinate of foot of perpendicular are 𝟏𝟐𝟐𝟗, −𝟏𝟖𝟐𝟗, 𝟐𝟒𝟐𝟗
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Chapter 11 Class 12 Three Dimensional Geometry
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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.