Question

Find the vector form of the equation of the plane that passes through the point P₀ = (1, - 2,3) and has normal vector n =<3,1,-1>.

Find an equation for the plane that contains the line x = - 1 +3t, y=5 +3t, z =2 +t and is parallel to the line of intersection of the planes x-2(y - 1) +3z =-1 and y =-2x - 1 =0

Answer 1

1)$x_o=1\\ y_o=-2\\ z_o=3\\ a=3\\ b=1\\ c=-1\\ a(x-x_o)+b(y-y_o)+c(z-z_o)=0\\ 3(x-1)+1(y-(-2))-1(z-3)=0\\ 3x+y-z+2=0\\ Let\ y=s,z=t\\ x=-\frac{2}{3}-\frac{1s}{3}+\frac{1t}{3}\\ y=0+1s+0t\\ z=0+0s+1t\\$

Vector form becomes

$<-\frac{2}{3},0,0>+s<-\frac{1}{3},1,0>+t<\frac{1}{3},0,1>\\$

2)

$x−2(y−1)+3z=−1\\ y = −2x −1\\ y=-2x-1\\ x+4x+4+3z=-1\\ y=-2x-1\\ z=-\dfrac{5}{3}x-\dfrac{5}{3}\\ <0,−1,−\frac{5}{3}>+t<1,−2,−\frac{5}{3}>\\ ​The\ plane\ contains\ the\ line\\ \dfrac{x-0}{1}=\dfrac{y+1}{-2}=\dfrac{z+\dfrac{5}{3}}{-\dfrac{5}{3}}\\ An\ equation\ for\ the\ plane\ parallel\ to\\ the\ line\ of\ intersection\ of\ the\ planes\\ ax+by+cz+d=0,ax+by+cz+d=0,\\ where\\ a-2b-\dfrac{5}{3}c=0\\ The\ plane\ contains\ the\ line\\ x = −1 + 3t, y = 5 + 3t, z = 2 + t\\ t=0: Point(-1, 5, 2)\\ t=-2: Point(-7, -1, 0)\\ -a+5b+2c+d=0\\ -7a-b+d=0\\ b=-7a+d\\ -a-35a+5d+2c+d=0\\ b=-7a+d\\ c=18a-3d\\ Substitute\\ a+14a-2d-30a+5d=0\\ d=5a\\ If\ a=1\\ d=5\\ b=-2\\ c=3\\ The\ equation\ for\ the\ plane\ is\\ x-2y+3z+5=0$