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- 1. 1a)Find parametric equations of the line of the intersection of the planes x-2yplus 7z=0 and 3xplus 5y-z=0 Solution Equation of the two planes: Plane 1: x - 2y + 7z = 0 and Plane 2: 3x + 5y - z = 0 normal vector of plane 1 is N1= (i - 2j + 7k) and of plane 2 is N2= (3i + 5j - k) let the intersection line be L and the direction vector of L be vector R ; now,since this line lies on both the planes it is perpendicular to both the normal vectors. therefore, R =N1 x N2 ..( x denotes cross product) or, R = -33i + 22j + 11k or, R = -3i + 2j + k now, we need a point which lies on this line(any point will do). that point will also lie on the two planes. we can see that (0,0,0) satisfies both the plane equations , therefore the line intersecting these 2 planes also passes through (0,0,0). parametric form of line : L = P + mR where P is a point on line L and R is the direction vector. therefore, L = m (-3i + 2j + k) , where m is a parameter. now if (Lx,Ly,Lz) is a point on this line and (Px,Py,Pz) is the another point lying on the same line.
- 2. then its co-ordinates are given as Lx = Px + m(-3) ; Ly = Py + m(2) ; Lz = pz + m(1) ; therfore, Lx = 0 - 3m or, Lx = -3m ; Ly = 0 + 2y or, Ly = 2m ; Lz = 0 + 1m or, Lz = m ;