Arectangular box is constructed in 3-space with one corner at the origin and other vertices at (1, 0, 0), (0, 4, 0), (0, 0, 2). find the length of the diagonal of the box.

The space diagonal will have length ...

... d = √(1² +4² +2²) = √(1 +16 +4) = √21

This can be found using the Pythagorean theorem. A drawing can help. Find the length of any face diagonal, then use that length as the leg of a right triangle whose hypotenuse is the space diagonal and whose other leg is the edge length not used in the first calculation.

if so 2130.12548...x^2

Step-by-step explanation: