Distance of point from plane
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Distance of point from plane
Last updated at April 16, 2024 by Teachoo
Question 10 If the points (1, 1 , p) and (β3 , 0, 1) be equidistant from the plane π β. (3π Μ + 4π Μ β 12π Μ) + 13 = 0, then find the value of p. The distance of a point with position vector π β from the plane π β.π β = d is |(π β.π β β π )/|π β | | Given, the points are The equation of plane is π β. (3π Μ + 4π Μ β 12π Μ) + 13 = 0 π β.(3π Μ + 4π Μ β 12π Μ) = β13 (1, 1, p) So, (π_1 ) β = 1π Μ + 1π Μ + pπ Μ (β3, 0, 1) So, (π_2 ) β = β3π Μ + 0π Μ + 1π Μ βπ β.(3π Μ + 4π Μ β 12π Μ) = 13 π β.(β3π Μ β 4π Μ + 12π Μ) = 13 Comparing with π β.π β = d, π β = β3π Μ β 4π Μ + 12π Μ d = 13 Magnitude of π β = β((β3)^2+(β4)^2+γ12γ^2 ) |π β | = β(9+16+144) = β169 = 13 Distance of point (ππ) β from plane |((π1) β"." π β" " β π)/|π β | | = |((1π Μ + 1π Μ + ππ Μ ).(β3π Μβ4π Μ+12π Μ )β13)/13| = |((1Γβ3)+(1Γβ4) +(πΓ12)β13)/13| = |(β3β4+12πβ13)/13| = |(12π β 20)/13| Distance of point (ππ) β from plane |((π2) β"." π β β π)/|π β | | = |((β3π Μ +0π Μ +1π Μ ).(β3π Μβ4π Μ+12π Μ )β13)/13| = |((β3Γβ3)+(0Γβ4) +(1Γ12)β13)/13| = |(9 + 0 +12β13)/13| = |8/13| = 8/13 Since the plane is equidistance from both the points, |(πππ β ππ)/ππ| = π/ππ |12πβ20| = 8 (12p β 20) = Β± 8 12p β 20 = 8 12p = 8 + 20 12p = 28 p = 28/12 p = 7/3 12p β 20 = β8 12p = β8 + 20 12p = 12 p = 12/12 p = 1 Answer does not match at end. If mistake, please comment