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Let OL, OM, ON be three mutually perpendicular edges of a rectangular box (cuboid) meeting at vertex O, with OL = 1 cm, OM = 2 cm, ON = 3 cm. Let l, m, n be the shortest distances between each of OL, OM, ON (respectively) and the body diagonal of the cuboid that does not intersect them. Find sqrt(13*l² + 10*m² + 5*n²).

1. sqrt(6)
2. sqrt(3)
3. 6/sqrt(14)
4. sqrt(6/7)

Correct answer: 6/sqrt(14)

Solution

Place O at origin, L at (1,0,0), M at (0,2,0), N at (0,0,3). The body diagonal from O to (1,2,3) shares vertex O with all three edges, so the skew body diagonal is the one from, say, (1,0,0) to (0,2,3) for OL. For OL (along x-axis from O): skew diagonal goes from (1,0,0) to (0,2,3), direction (-1,2,3). Line 1: point O=(0,0,0), dir (1,0,0). Line 2: point (1,0,0), dir (-1,2,3). Connecting vector: (1,0,0)-(0,0,0) = (1,0,0). Cross product (1,0,0)x(-1,2,3) = (0*3-0*2, 0*(-1)-1*3, 1*2-0*(-1)) = (0,-3,2). |cross| = sqrt(0+9+4) = sqrt(13). l = |(1,0,0).(0,-3,2)|/sqrt(13) = |0|/sqrt(13) = 0. Hmm, they share a point. Let me reconsider: the body diagonal skew to OL must not intersect OL. The four body diagonals are: O to (1,2,3), (1,0,0) to (0,2,3), (0,2,0) to (1,0,3), (0,0,3) to (1,2,0). OL is from O=(0,0,0) to (1,0,0). The diagonal (1,0,0) to (0,2,3) shares point (1,0,0) with OL, so they intersect. The diagonal (0,2,0) to (1,0,3): does it intersect OL? OL: (t,0,0). This diagonal: (2s, 2-2s, 3s) for the other parametrization... Let me use the correct diagonal. Diagonal from (0,2,0) to (1,0,3): direction (1,-2,3). For intersection with OL (t,0,0): need 0+s*1=t, 2+s*(-2)=0, 0+s*3=0. From 3rd: s=0, from 2nd: 2=0, contradiction — skew. l = |((0,2,0)-(0,0,0)).(1,0,0)x(1,-2,3)| / |(1,0,0)x(1,-2,3)|. (1,0,0)x(1,-2,3) = (0*3-0*(-2), 0*1-1*3, 1*(-2)-0*1) = (0,-3,-2). |(0,-3,-2)| = sqrt(13). Dot (0,2,0).(0,-3,-2) = -6. l = 6/sqrt(13). Similarly for OM and ON.

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