WebThe equation of plane through the line of intersection of the planes `2x+3y+4z-7=0,x+y+z-1=0` and Doubtnut 2.72M subscribers Subscribe 428 views 2 years ago The equation of plane through... WebMar 30, 2024 · Ex 11.3, 9 Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1). Equation of a plane passing through the intersection of planes A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 and through the point (x1,
1. Find the distance from the point (1, 4, 1) to the plane 2x y z …
WebAnswered: 7. Find the line tangent to the… bartleby. ASK AN EXPERT. Math Advanced Math 7. Find the line tangent to the intersection of the surfaces and at the point (√2, 1, -2). x² + … WebDec 9, 2015 · How to find the point where the line x = –1 – t, y = 2 + t, z = 1+ t intersects the plane 3x + y + 3z = 1 ? Calculus Parametric Functions Introduction to Parametric Equations 1 Answer Alan P. Dec 9, 2015 (x,y,z) = (0,1,0) Explanation: Given [1] XXXx = −1 −t [2] XXXy = 2 + t [3] XXXz = 1 + t [4] XXX3x + y + 3z = 1 traffic 11235
The distance from the origin to the plane x+3y-2z+1=0 is
WebAnswered: 7. Find the line tangent to the… bartleby. ASK AN EXPERT. Math Advanced Math 7. Find the line tangent to the intersection of the surfaces and at the point (√2, 1, -2). x² + y² +2²-3y + 2z = 0 x² - y²-2x = 5. 7. Find the line tangent to the intersection of the surfaces and at the point (√2, 1, -2). x² + y² +2²-3y + 2z ... WebView the full answer. Transcribed image text: Find a set of scalar parametric equations for the line formed by the two intersecting planes. 3x + 3y + 2z + 1 = 0 4x - 3y + 2z-4 = 0 a) ©$ (1) = 1 + 41, y (0) = + „v, 260) = 5-1 - 71 b) X (t) = 41, y (t) = 5 + 6 77 34, 2 (1) 3 77 c) ©* (0) = 41, y (t) = + 1, 20) = d) x () = 41, y () == + 1, 20 ... WebSolution Verified by Toppr Correct option is A) Equation of line parallel to x+y=1 is x+y+c=0 Let it passes through (1,1) so the line is along direction of x+y=1 ⇒1+1+c=0 ⇒c=−2 ⇒x+y−2=0.....(i) Given line is 2x−3y−4=0....(ii) Solving (i) and (ii) ⇒x=2,y=0 So the point of intersection is (2,0) So the distance of (1,1) from 2x−3y−4=0 along x+y+1=0 is traffic100k review