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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


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 (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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.