In the figure, a cube of edge length a = 3.80 m sits with one corner at the orig

Question

In the figure, a cube of edge length a = 3.80 m sits with one corner at the origin of an xyz coordinate system. A body diagonal is a line that extends from one corner to another through the center. In unit-vector notation, what is the body diagonal that extends from the corner at (a) coordinates (0, 0, 0), (b) coordinates (a, 0, 0), (c) coordinates (0, a, 0), and (d) coordinates (a, a, 0)? (e) Determine the angles that the body diagonals make with the adjacent edges. (f) Determine the length of the body diagonals.

Explanation / Answer

a) As can be seen from the figure, the point diametrically opposite the origin (0,0,0) has position
vector aˆi+ aˆj + aˆk and this is the vector along the “body diagonal”.

{ai,aj,ak}

b) From the point (a,0,0) which corresponds to the position vector aˆi, the diametrically opposite point is (0,a,a) with position vector aˆj + aˆk. Thus, the vector along the line is the difference aˆi + aˆj + aˆk.

{-ai, aj, ak}

c) If the starting point is (0,a,0) with the corresponding position vector aˆj, the diametrically opposite point is (a,0,a) with position vector aˆi + aˆk. Thus, the vector along the lines is the difference   aˆi aˆj + aˆk.

{ai, -aj, ak}

d)If the starting point is (a,a,0) with the corresponding position vector aˆi + aˆj, the diametrically opposite point is (0,0,a) with the position vector aˆk. Thus, the vector along the line is the difference aˆi aˆj + aˆk.

{-ai, -aj, ak}

(e) = cos^-1(1/3) 54.73561° 54.7°

(f) The length of the diagonals is given by: sqrt(a2 + a2 + a2 ) = a3

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