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Example 16 - Find coordinates of foot of perpendicular from - Examples

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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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 𝑎﷮1﷯﷮ 𝑎﷮2﷯﷯ = 𝑏﷮1﷯﷮ 𝑏﷮2﷯﷯ = 𝑐﷮1﷯﷮ 𝑐﷮2﷯﷯ = k 𝑥﷮1﷯﷮2﷯ = 𝑦﷮1﷯﷮ − 3﷯ = 𝑧﷮1﷯﷮4﷯ = 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 = 6﷮29﷯ So, 𝑥﷮1﷯ = 2k = 2 × 6﷮29﷯ = 12﷮29﷯ 𝑦﷮1﷯ = –3k = −3 × 6 ﷮29﷯ = −18﷮29﷯ & 𝑧﷮1﷯ = 4k = 4 × 6﷮29﷯ = 24﷮29﷯ Therefore, coordinate of foot of perpendicular are 𝟏𝟐﷮𝟐𝟗﷯, −𝟏𝟖﷮𝟐𝟗﷯, 𝟐𝟒﷮𝟐𝟗﷯﷯

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