Question #211900

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.


1
Expert's answer
2021-06-30T16:10:01-0400

1.1. The plane x+2y+3z9=0x+2y+3z-9=0 is orthogonal to the vector n=(1,2,3)\vec{n}=(1,2,3).

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

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

But since v,n=11+(1)2+(2)3=80\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 4xy+2z+1=04x-y+2z+1=0 is orthogonal to the vector n=(4,1,2)\vec{n}=(4,-1,2).

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

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

But since v,n=34+2(1)+(1)2=80\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+3tx(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).

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

x(t)=1+6t,y(t)=2,z(t)=3tx(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.


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