Example 16 - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
Last updated at Feb. 1, 2020 by Teachoo
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Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
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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)
