Question

(5) Equations for Planes. (a) Find an equation of the plane passing through (1,2,3) that is parallel to the plane r -y + 2z = 5. (b) Find an equation of the plane passing through the point (0,1,0) and containing the line r = (-t, 2t, 4t).

Answer 1

Detailed solution is attached here.

la) Solection=) The general foom g the equation of plane paraller to line ax+ by + cz + d 20 18 giren by - 4Xby+CZ+d+2=0 LEI Where a 18. determined by puthing the value of point to which the plane 18 passing througho Hence equation of plane parallel to plane X-y + 2z = 5 1'8, X -4+ 27 +9 = 5 a As the above plane is passing through the point (1,2,3). therefore - B 1 - 2 + 6 +2=5 de Mething value of a cm equation Lise s Se-Y.+22=5] 16) The general form of equation for a plane, where CCO , Yo, Zo) ve any point on the plane and (a, b, c) is the normal vector of the plane. Q (QC-X09 + 6 (-) 40 (2- Zo)=0 - (2) The plane 1'8 passing through the point (0,2,0) hence (xo, Yo, Z) = (0,1,0)

* To find the normal vector(0,b,c), we need to vectors in the plane · Then we can find normal vector by their cross product * The plane contains the line ř=-, 27, 47). Therefore One vector in plane i'd directional vector of ñ tes given by covered ) = {1,2,4) * gt 18 given that (0,1,0) is a point lies on the plane. We can find another point on the plane by Peething any random value of t in equation g line passing through the plane. Substituting t=0 Crves (0,0,0) 98 another pornt ich the plane. Vector from point (0, 1, 0) to (0,0,0) 08 <0-0, 0-1,0-02= (0,-1,0> * Hence Normal vector can be obtained by cross Product of vector 4-2, 2, 4) and vector (0,-1,0) Normal vector = (-1,2,4><0,-1,0 > = 54,0,1)

Peetting the value of normal vector <4,0,1% =20,6,c> and the point to,1,0) = (xo, Yo, Zo) in equation (2) 4(2-0) +019-1) + 1 (2-0)=0 | 4X+z=0