   1. Chapter 11 Class 12 Three Dimensional Geometry
2. Serial order wise
3. Examples

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 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﷯﷯

Examples 