Math 225: Calculus III    Quiz 1    1/21&23/97

<<Calculus`Math225`

i = {1,0,0}; j = {0,1,0}; k = {0,0,1};

1) Determine the (simplified) equation of the plane whose points are equidistant from the points {1,2,3} and {6,7,8}.

Solution

If {x,y,z} is a point in the plane, then the distance of this point from the two given points is

Sqrt[(x-1)^2 + (y-2)^2 + (z-3)^2] ==
    Sqrt[(x-6)^2 + (y-7)^2 + (z-8)^2]

(x-1)^2 + (y-2)^2 + (z-3)^2 ==
    (x-6)^2 + (y-7)^2 + (z-8)^2 //Simplify

Square both sides and expand everything:

Cancel terms:

14 - 2x - 4y - 6z == 149 - 12x - 14y - 16z

Collect like terms:

10x + 10y + 10z == 135

or

x + y + z == 13.5

2) Find a vector of the form {1,y,1} that forms a 60 degree angle with the x-axis.

Solution

The direction of the x-axis is represented by the unit vector i = {1,0,0}. The dot product of two vectors equals the product of their lengths times the cosine of the angle between them.

{1,y,1}.i == Norm[{1,y,1}]Norm[i] Cos[Pi/3]

Solve[%]

"By hand": Square both sides and simplify:
      4 == 2 + y^2
    y^2 == 2
      y == ±Sqrt[2]
There are two points that qualify: {1,-Sqrt[2],1} and {1,Sqrt[2],1}.

Here's a picture (not required):

v1 = Vector[{0,0,0},{1,-Sqrt[2],1}];
v2 = Vector[{0,0,0},{1, Sqrt[2],1}];
w  = Vector[{0,0,0},{1,0,0}];

Show[v1,v2,w, ViewPoint->{3.086, 0.840, 1.104}, Axes->True]