Example 16
Last updated at Dec. 8, 2016 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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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.