Angles of 3D figure---Using Vectors

Question

A rectangular box has length 15 inches, width 7 inches, and a height of 16 inches. Find the angle between the diagonal of the box and the diagonal of its base. The angle should be measured in radians.

I got 0.7805624981 radians, but it says this is wrong.

If you could explain how you get your answer, that would be helpful!

Answer 1

Consider the base of the box in the x-y plane.

Let one corner of the box be located at (0,0,0).

Moving along the x-axis for a length of 15 takes you to (15,0,0).

Along the y-axis for a width of 7-> (15,7,0).

The vector from the origin to that far corner is <15,7,0> which is the diagonal of the base.

From that far corner move up 16 in the z direction for the height -> (15,7, 16).

The vector from the origin to that point is <15,7, 16>. This represents the diagonal of the box.

We want to find the angle between the vectors v₁=<15,7,0> and v₂=<15,7,16>.

cosθ= (v₁dot v₂)/(|v₁||v₂|)

The dot product is: (15)(15) + (7)(7) + (0)(16) = 274

Magnitude of v₁= √(15²+7²+0²) = √274

Magnitude of v₂= √(15²+7²+16²) = √530

cosθ = (274)/[√274 √530] ≈ 0.719013999916

θ = cos^-1(0.719013999916) ≈ 0.768413765187

This Desmos 3D graphs helps picture the diagonals:

desmos.com/3d/h7jlmgrkq9