Answer to Question #211900 in Analytic Geometry for Faith

Question

1.determine whether the given line and the given planes are parallel:

1.1.x=1+t,y=-1-t,z=-2t and x+2y+3z-9=0.

1.2.<0,1,2>+t<3,2,-1> and 4x-y+2z+1=0.

2.

2.1.find parametric equations of the line that passes through the point p=(2,0,-1) and is parallel to the vector n=<2,1,3>.

2.2.find parametric equations of the line that passes through the points a=(1,2,-3) and b=(7,2,-4).

2.3.find parametric equations for the line intersection of the planes -5x+y-2z=3 and 2x-3y+5z=-7.

Answer 1

1.1. The plane $x+2y+3z-9=0$ is orthogonal to the vector $\vec{n}=(1,2,3)$.

The line $x=1+t,\,y=-1-t,\,z=-2t$ has a direction vector $\vec{v}=(1,-1,-2)$.

It is parallel to the plane with a normal vector $\vec{n}=(1,2,3)$ , if and only if $\langle \vec{v}, \vec{n}\rangle=0$.

But since $\langle \vec{v}, \vec{n}\rangle = 1\cdot 1+(-1)\cdot 2+(-2)\cdot 3=-8\ne 0$, this line and this plane are not parallel.

1.2. The plane $4x-y+2z+1=0$ is orthogonal to the vector $\vec{n}=(4,-1,2)$.

The line $<0,1,2>+t<3,2,-1>$ has a direction vector $\vec{v}=(3,2,-1)$.

It is parallel to the plane with a normal vector $\vec{n}=(4,-1,2)$ , if and only if $\langle \vec{v}, \vec{n}\rangle=0$.

But since $\langle \vec{v}, \vec{n}\rangle = 3\cdot 4+2\cdot (-1)+(-1)\cdot 2=8\ne 0$, this line and this plane are not parallel.

2.1.find parametric equations of the line that passes through the point p=(2,0,-1) and is parallel to the vector n=<2,1,3>.

$x(t)=2+2t,\, y(t)=t,\, z(t) = -1+3t$

2.2.find parametric equations of the line that passes through the points a=(1,2,-3) and b=(7,2,-4).

$\vec{ab}=(7,2,-4)-(1,2,-3)=(6,0,-1)$

$x(t)=1+6t,\, y(t)=2,\, z(t) = -3-t$

2.3.find parametric equations for the line intersection of the planes -5x+y-2z=3 and 2x-3y+5z=-7.

Let z=t be a parameter. Then

-5x+y=3+2z=3+2t, y=5x+3+2t

2x-3y=-5z-7, 2x-(5x+3+2t)=-5t-7

-3x=-3t-4

x=t+4/3, y=5(t+4/3)+3+2t = 7t+29/3, z=t. This is parametric equation for the line intersection of the given planes.