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Example 16 - Find coordinates of foot of perpendicular from

Example 16 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example 16 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Example 16 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

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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 (𝑢𝑷) βƒ— is parallel to normal vector 𝒏 βƒ— Given equation of the plane is 2x βˆ’ 3y + 4z βˆ’ 6 = 0 2x βˆ’ 3y + 4z = 6 So, Normal vector = 𝒏 βƒ— = 2π’Š Μ‚ – 3𝒋 Μ‚ + 4π’Œ Μ‚ Since, (𝑢𝑷) βƒ— and 𝒏 βƒ— are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional π‘Ž_1/π‘Ž_2 = 𝑏_1/𝑏_2 = 𝑐_1/𝑐_2 = k π‘₯_1/2 = 𝑦_1/( βˆ’ 3) = 𝑧_1/4 = k x1 = 2k , y1 = βˆ’3k , z1 = 4k (𝑢𝑷) βƒ— = x1π’Š Μ‚ + y1𝒋 Μ‚ + z1π’Œ Μ‚ Direction ratios = x1, y1, z1 ∴ a1 = x1 , b1 = y1, c1 = z1 𝒏 βƒ— = 2π’Š Μ‚ – 3𝒋 Μ‚ + 4π’Œ Μ‚ Direction ratios = 2, βˆ’3, 4 ∴ a2 = 2 , b2 = –3, c2 = 4 Also, point P(x1, y1, z1) lies in the plane. Putting P (2k, βˆ’ 3k, 4k) in equation of plane 2x βˆ’ 3y + 4z = 6 2(2k) βˆ’ 3(βˆ’3k) + 4(4k) = 6 4k + 9k + 16k = 6 29k = 6 ∴ k = 6/29 So, π‘₯_1 = 2k = 2 Γ— 6/29 = 12/29 𝑦_1 = –3k = βˆ’3 Γ— (6 )/29 = (βˆ’18)/29 𝑧_1 = 4k = 4 Γ— 6/29 = 24/29 Therefore, coordinate of foot of perpendicular are (𝟏𝟐/πŸπŸ—, ( βˆ’πŸπŸ–)/πŸπŸ—,πŸπŸ’/πŸπŸ—)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.