Past exam of the mathematics course of the University of Cambridge 2005 ia Paper 1 7A Solution Created 2026-09-24 Updated 2026-10-14
For , define the orthogonal projection of onto its span byThe first is parallel or antiparallel to , or zero, and . This proves the decomposition. If as well, comparison with yieldsIndeed shows that the ratio lies in . There is a unique with this cosine. The definition is symmetric in the two vectors, unchanged by orthogonal coordinate changes and positive rescaling, and agrees with ordinary plane geometry in their span: projecting a unit vector onto the other gives its cosine. Parallel, perpendicular and antiparallel directions give respectively. The zero vector has no defined angle; if , the nonzero-axis projection formula itself must not be used.
The vertices of the centered hypercube are sign vectors . A body diagonal identifies and , hence there are such diagonals. If two sign vectors disagree in positions, their dot product is . For odd this integer is odd and cannot be zero. Thus no two body diagonals are perpendicular in odd dimension, by the orthogonality of hypercube body diagonals.
In four dimensions, mutually orthogonal nonzero vectors are linearly independent, so at most four mutually perpendicular diagonals are possible. Four are achieved by the directionsEach has squared norm four and every distinct pair has zero dot product; these are the rows of a Hadamard matrix. Therefore the maximum is four.
For distinct diagonals, the Hamming distance is ; would represent the same line. Oriented vertex directions therefore give cosines . The angles areFor unoriented diagonals, the usual smaller angle between the lines is or ; the supplementary intersection angle also occurs when the two rays are chosen oppositely. Stating this convention avoids counting a line and its reverse as different diagonals.