# A gallery is about to open an exhibit in a room that is 22 ft long, 24 ft wide and 16 ft high. the edge of the door to the exhibit is is on the 22 ft side, 4 ft from 1 wall. For special effects...

A gallery is about to open an exhibit in a room that is 22 ft long, 24 ft wide and 16 ft high. the edge of the door to the exhibit is is on the 22 ft side, 4 ft from 1 wall. For special effects lightning as a guest enters the door, let * u* be the vector representing the distance from the edge of the door (at the floor) to the farthest bottom corner of the room, and let

*be the vector from the same point to the farthest upper corner*

**v**Find |v|, the angle between * u* and

*using cos theta = u.v/|u||v|. Verify result using right triangle trigonometry.*

**v**

*print*Print*list*Cite

A gallery is about to open an exhibit in a room that is 22 ft long, 24 ft wide and 16 ft high. The edge of the door to the exhibit is is on the side that has a length of 22 ft and is 4 ft from the wall.

`vec u` represents the vector from the edge of the door (at the floor) to the farthest bottom corner of the room, and `vec v` is the vector from the same point to the farthest upper corner.

The magnitude of `vec v` is `sqrt(18^2+24^2+16^2)` = 34

The magnitude of `vec u` is `sqrt(18^2+24^2)` = 30

If the edge of the door is taken as (0,0,0) , `vec u` = [24,18,0] and `vec v` = [24,18,16]

`vec u @ vec v` = 24*24 + 18*18 + 16*0 = 900

`cos theta = 900/(30*34)`

=> `theta = cos^-1(900/(30*34))` = 28.072 degrees

The angle between the vectors `vec u` and `vec v` using the right triangle formed is : `tan^-1(16/30)` `~~` 28.07 degrees.