Webb4 jan. 2024 · Find the equation of the plane that passes through the points. The answer writes: Let x = P 1 P 2 → = [ 1 2 1] and y = P 1 P 3 → = [ 2 − 4 1] The normal vector N … WebbEquation of any plane passing through (-2, -2, 2) is a (x + 2) + b (y + 2) + c (z - 2) = 0 since it contains the line joining the points (1, 1, 1) and (1, -1, 2), it contains these points as well so that 3a + 3b - c = 0 and 3a + b + 0 = 0 on solving we get 1a = −3b = −6c and thus the equation of the plane is x+2− 3(y+ 2)− 6(z −2) = 0 or x−3y −6z …

Distance Between Point and Plane - Formula, Proof, Examples

WebbQuestion Equation of the plane through the points (2,−1,0) and (3,−4,5) parallel to the line 2x= 3y=4z is given by: A 29x−27y−22z−85=0 B 29x+27y−22z−85=0 C 29x−27y+22z−85=0 D None of these Medium Solution Verified by Toppr Correct option is A) Was this answer helpful? 0 0 Similar questions Webb13 apr. 2024 · Plane P3 is passing through (1,1,1) and line of intersection of P1 and P2 where P1: 2x - y + z = 5 and P2: x + 3y + 2z + 2 = 0. Then distance of ... $ and Passing … imphal airport india

Equation of the Plane Passing Through Two Points and is

Webb11 feb. 2024 · closed Feb 17 by SukanyaYadav If the equation of the plane passing through the point (1,1,2) and perpendicular to the line x - 3y + 2z - 1 = 0 , 4x - y + z is Ax + By + Cz = 1, then 140 (C - B + A) is equal to _____. jee main 2024 1 Answer +1 vote answered Feb 11 by Rishendra (52.8k points) selected Feb 17 by SukanyaYadav Best answer WebbThe vector equation of the plane passing through the intersection of the planes → r ⋅(^i +^j +^k) =1 and → r ⋅(^i −2^j)= −2, and the point (1,0,2) is A → r ⋅(^i −7^j +3^k)= 7 3 B → r ⋅(^i +7^j +3^k)=7 C → r ⋅(3^i +7^j +3^k)=7 D → r ⋅(^i +7^j +3^k)= 7 3 Solution The correct option is C → r ⋅(^i +7^j +3^k)= 7 Webb12 nov. 2024 · Introduction Equation of the Plane Passing Through Two Points and is Perpendicular to a Given Plane Glass of Numbers 4.48K subscribers Subscribe 203 Share 11K views 2 years ago … imphal airport project