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Example 16 - Chapter 11 Class 12 Three Dimensional Geometry
Last updated at Feb. 1, 2020 by Teachoo
Transcript
Example 16 Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x β 3y + 4z β 6 = 0. Let point P(x1, y1, z1) be foot of perpendicular from origin Since perpendicular to plane is parallel to normal vector Vector (πΆπ·) β is parallel to normal vector π β Given equation of the plane is 2x β 3y + 4z β 6 = 0 2x β 3y + 4z = 6 So, Normal vector = π β = 2π Μ β 3π Μ + 4π Μ Since, (πΆπ·) β and π β are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional π_1/π_2 = π_1/π_2 = π_1/π_2 = k π₯_1/2 = π¦_1/( β 3) = π§_1/4 = k x1 = 2k , y1 = β3k , z1 = 4k (πΆπ·) β = x1π Μ + y1π Μ + z1π Μ Direction ratios = x1, y1, z1 β΄ a1 = x1 , b1 = y1, c1 = z1 π β = 2π Μ β 3π Μ + 4π Μ Direction ratios = 2, β3, 4 β΄ a2 = 2 , b2 = β3, c2 = 4 Also, point P(x1, y1, z1) lies in the plane. Putting P (2k, β 3k, 4k) in equation of plane 2x β 3y + 4z = 6 2(2k) β 3(β3k) + 4(4k) = 6 4k + 9k + 16k = 6 29k = 6 β΄ k = 6/29 So, π₯_1 = 2k = 2 Γ 6/29 = 12/29 π¦_1 = β3k = β3 Γ (6 )/29 = (β18)/29 π§_1 = 4k = 4 Γ 6/29 = 24/29 Therefore, coordinate of foot of perpendicular are (ππ/ππ, ( βππ)/ππ,ππ/ππ)
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Chapter 11 Class 12 Three Dimensional Geometry
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.