Initializations

Arrow3D[{x1_,y1_,z1_},{x2_,y2_,z2_},color_]:=Module[{line,head,headsize},
headsize=0.015*(Sqrt[(((x1-x2))^2)+(((y1-y2))^2)+(((z1-z2))^2)]);
head=Sphere[{x2,y2,z2},headsize];
line=Line[{{x1,y1,z1},{x2,y2,z2}}];
Return[Graphics3D[{color,Thick,line,head}]];
]

Lines and Planes in Space

Example 12.5.22

We want to find an equation for the plane that passes through (1,-1,3) and is parallel to the plane 3x+y+z=7.

Solution

A plane through a point ((Subscript[x, 0]),(Subscript[y, 0]),(Subscript[z, 0])) normal to the vector A i+B j+C k is given by A(x-(Subscript[x, 0]))+B(y-(Subscript[y, 0]))+C(z-(Subscript[z, 0]))=0.

The plane we require will be parallel to the plane 3x+y+z=7, which has a normal given by 3 i+1 j+1 k

The plane we are looking for is therefore given by:

A(x-(Subscript[x, 0]))+B(y-(Subscript[y, 0]))+C(z-(Subscript[z, 0]))=0
3(x-1)+1(y+1)+1(z-3)=0

Visual Check of Solution

Here's a sketch of what we have, the plane and the normal to the plane:

surface1 = -3x-y+7;
surface1plot =Plot3D[surface1,{x,-5,5},{y,-5,5}, Axes->True,AxesLabel->{"x-axis","y-axis","z-axis"},PlotStyle->Opacity[0.75],Mesh->False,AspectRatio->1];

normalvector={3,1,1}
plot2=Show[surface1plot,Arrow3D[{0,0,0},normalvector,Red],PlotRange->{{-5,5},{-5,5},{-5,5}}]

Here is a sketch of the situation, with a sphere drawn at the point (1,-1,3)

surface2 = 3(x-1)+1(y+1)+1(z-3);
surface2plot = ContourPlot3D[surface2,{x,-5,5},{y,-5,5},{z,-5,5},Contours->0, Axes->True,AxesLabel->{"x-axis","y-axis","z-axis"},Mesh->False];

Show[plot2,surface2plot,Graphics3D[Sphere[{1,-1,3},0.3]],AspectRatio->1]

Example 12.5.30

Find a plane that is determined by the following two lines:
L1: x=t, y=3-3t, z=-2-t
L2: x=1+s,y=4+s,z=-1+s

Solution

To determine a plane, we need to know a point on the plane and a normal vector to the plane.

Point on the plane is found from any point on the given lines. 
Choose L1with t=0 to get the point (0,3,-2)as a point on the plane.

The normal vector to the plane can be found by taking the cross product of the two vectors that come for the lines.
L1 has two points on it:
t = 0: (0,3,-2)  and
t=1: (1,0,-3). 
A vector along L1 is (Subscript[v, 1])=<1,-3,-1>.

L2 has two points on it:
s= 0: (1,4,-1)  and
s=1: (2,5,0). 
A vector along L2 is (Subscript[v, 2])=<1,1,1>.

Let's work out the cross product using Mathematica, but you should be able to do this by hand

Needs["VectorAnalysis`"]

CrossProduct[{1,-3,-1},{1,1,1}]

Check:

{{i,j,k},{1,-3,-1},{1,1,1}}
MatrixForm[%]
Det[%%]

So the equation of the plane is given by

A(x-(Subscript[x, 0]))+B(y-(Subscript[y, 0]))+C(z-(Subscript[z, 0]))=0
-2(x-0)-2(y-3)+4(z+2)=0
-2x-2(y-3)+4(z+2)=0

Visual Check of Solution

Solve[-2x-2(y-3)+4(z+2)==0,z]

surface1 =(1/2) (-7+x+y)
surface1plot =Plot3D[surface1,{x,-15,15},{y,-15,15}, Axes->True,AxesLabel->{"x-axis","y-axis","z-axis"},Mesh->False,PlotStyle->Opacity[0.75]];

normalvector={-2,-2,4}
plot2=Show[surface1plot,Arrow3D[{0,0,0},-1*normalvector,Red],Graphics3D[Sphere[{0,3,-2},1]],AspectRatio->1]

Example 12.5.61

Determine if the following two lines are parallel, skew, or intersect. If they intersect, find the point of intersection.
L1: x=3+2t,y=-1+4t,z=2-t
L2: x=1+4s, y=1+2s, z=-3+4s

Solution

I'll do one set of the lines here, and leave the others for you to verify. Let's determine if L1 and L2 are parallel, skew, or intersect::

L1: x=3+2t,y=-1+4t,z=2-t
L2: x=1+4s, y=1+2s, z=-3+4s

L1 goes through (3,-1,2) and (5,3,1) and therefore has a direction vector <-2,-4,1>.

L2 goes through (1,1,-3) and (-3,-1,-7) and therefore has a direction vector <4,2,4>.

Since the two drection vectors are not parallel, the lines are not parallel.

For the two lines to intersect, the coordinates must be same for the same:
x: 3+2t=1+4s
y: -1+4t=1+2s
z: 2-t=-3+4s

Pick the first two equations, and solve for s and t.

3+2t=1+4s
 -1+4t=1+2s

Solve[{3+2t==1+4s,-1+4t==1+2s},{s,t}]

Now, if these values also make the third equation, 2-t=-3+4s, true, then the two lines intersect. We can see that s=t=1does make this equation true, so the lines L1 and L2 do intersect when s=t=1, which occurs at (5,3,1).

Visual check of solution

L1=ParametricPlot3D[{3+2t,-1+4t,2-t},{t,-3,3},PlotStyle->{Red,Thick}];
L2=ParametricPlot3D[{1+4s,1+2s,-3+4s},{s,-3,3},PlotStyle->{Blue,Thick}];
intersection={5,3,1};
Show[L1,L2,Graphics3D[Sphere[intersection,0.5]],AspectRatio->1]

Example 12.5.64

Use the component form to generate an equation for the plane through (4,1,5)  and normal to <1,-2,1>. Then, generate another equation for the same plane using the point (3,-2,0) and the normal to <-(Sqrt[2]),2(Sqrt[2]),-(Sqrt[2])>.

Solution

A plane through a point ((Subscript[x, 0]),(Subscript[y, 0]),(Subscript[z, 0])) normal to the vector A i+B j+C k is given by A(x-(Subscript[x, 0]))+B(y-(Subscript[y, 0]))+C(z-(Subscript[z, 0]))=0.

Form 1:

A(x-(Subscript[x, 0]))+B(y-(Subscript[y, 0]))+C(z-(Subscript[z, 0]))=0
1(x-4)-2(y-1)+1(z-5)=0

Form 2:

A(x-(Subscript[x, 0]))+B(y-(Subscript[y, 0]))+C(z-(Subscript[z, 0]))=0
-(Sqrt[2])(x-3)+2(Sqrt[2])(y+2)-(Sqrt[2])(z-0)=0
-(Sqrt[2])(x-3)+2(Sqrt[2])(y+2)-(Sqrt[2])z=0

Visual Check of Solution

Solve[1(x-4)-2(y-1)+1(z-5)==0,z]

surface1 =7-x+2 y
surface1plot =Plot3D[surface1,{x,-15,15},{y,-15,15}, Axes->True,AxesLabel->{"x-axis","y-axis","z-axis"},Mesh->False,PlotStyle->Opacity[0.75]];

normalvector={1,-2,1}
plot2=Show[surface1plot,Arrow3D[{0,0,0},3*normalvector,Red],Graphics3D[Sphere[{4,1,5},1]],AspectRatio->1,PlotRange->{{-15,15},{-15,15},{-15,15}}]

Solve[-(Sqrt[2])(x-3)+2(Sqrt[2])(y+2)-(Sqrt[2])z==0,z]

surface2=-((-7 (Sqrt[2])+(Sqrt[2]) x-2 (Sqrt[2]) y)/(Sqrt[2]))
surface2plot =Plot3D[surface2,{x,-15,15},{y,-15,15}, Axes->True,AxesLabel->{"x-axis","y-axis","z-axis"},Mesh->False,PlotStyle->Opacity[0.75]];

normalvector={-(Sqrt[2]),2(Sqrt[2]),-(Sqrt[2])}
plot3=Show[surface2plot,Arrow3D[{0,0,0},5*normalvector,Red],Graphics3D[Sphere[{3,-2,0},1]],AspectRatio->1,PlotRange->{{-15,15},{-15,15},{-15,15}}]

Show[plot2,plot3]

Example 12.5.70

Find a plane through the origin that meets the plane M: 2x+3y+z=12 in a right angle. How do you know your plane is perpendicular to M?

Solution

If the plane we seek meets the plane M: 2x+3y+z=12 at a right angle, then it will be at right angles to any line in the plane M.

A plane through a point ((Subscript[x, 0]),(Subscript[y, 0]),(Subscript[z, 0])) normal to the vector A i+B j+C k is given by A(x-(Subscript[x, 0]))+B(y-(Subscript[y, 0]))+C(z-(Subscript[z, 0]))=0.

Since the plane we seek will go through the origin, all we need is to find a vector that lies on the plane M.

Solve[2x+3y+z==12,z]

Two points on the plane M are (0,0,12) and (1,1,7). A vector in the plane M is therefore <-1,-1,5>. The plane we seek is given by
A(x-(Subscript[x, 0]))+B(y-(Subscript[y, 0]))+C(z-(Subscript[z, 0]))=0
-1(x-0)-1(y-0)+5(z-0)=0
-x-y+5z=0

Visual Check of Solution

Solve[2x+3y+z==12,z]

surface1 =12-2 x-3 y
surface1plot =Plot3D[surface1,{x,-15,15},{y,-15,15}, Axes->True,AxesLabel->{"x-axis","y-axis","z-axis"},Mesh->False,PlotStyle->Opacity[0.75]];

surface2 =(x+y)/5
surface2plot =Plot3D[surface2,{x,-15,15},{y,-15,15}, Axes->True,AxesLabel->{"x-axis","y-axis","z-axis"},Mesh->False,PlotStyle->Opacity[0.75]];

plot2=Show[surface1plot,surface2plot,
Graphics3D[{Red,Thick,Line[{{0,0,12},{1,1,7}}]}],
Graphics3D[Sphere[{0,0,12},1]],
Graphics3D[Sphere[{1,1,7},1]],
Graphics3D[Sphere[{0,0,0},1]],
AspectRatio->1,PlotRange->{{-15,15},{-15,15},{-15,15}}]