Thread: Advanced Calculus 1 question

Here is the question from Multivariable Calculus by James Stewart 6th Edition:

"Find the angle between a diagonal of a cube and a diagonal of one of its faces." This is no differentiation or integration question. It is under the dot product section. I think you need to solve it through direction angles and direction cosines.

I drew a cube with one of its corner points on the origin of the x, y, and z-axis. The diagonal of the cube can be seen as the diameter of a sphere and therefore is 2r (r=radius). What can the length of the diagonal of one of the faces of the cube be? I don't think its half that of the diagonal of the whole cube. But if it is then it would seem one line is perpendicular to the other and the angle would be PI/2 in radians.

Originally Posted by Undefdisfigure

Here is the question from Multivariable Calculus by James Stewart 6th Edition:

"Find the angle between a diagonal of a cube and a diagonal of one of its faces." This is no differentiation or integration question. It is under the dot product section. I think you need to solve it through direction angles and direction cosines.

I drew a cube with one of its corner points on the origin of the x, y, and z-axis. The diagonal of the cube can be seen as the diameter of a sphere and therefore is 2r (r=radius). What can the length of the diagonal of one of the faces of the cube be? I don't think its half that of the diagonal of the whole cube. But if it is then it would seem one line is perpendicular to the other and the angle would be PI/2 in radians.

Take one corner as origin and the others as (0,0,1), (0,1,0), ...

The diagonal of the cube starting at the origin is then the vector d1=(1,1,1),
and a face diagonal starting from the same point is d2=(1,1,0).

Now the dot product of these is:

|d1| |d2| cos(theta) = (1,1,1).(1,1,0) = 2.

so:

theta= arccos(2/[|d1| |d2|])

RonL