Introduction to Quaternions

WE have thought it would be acceptable to many students if we should give as an Appendix a brief, and in some cases even a detailed, solution of the most important and most difficult of the ADDITIONAL EXAMPLES. In doing so, we would add as a word of advice, that our solutions be employed simply for the purpose of comparison with those which shall occur to the student himself.

Chap. II.

EX. 4. If

  AB = α, BC = β, AP = mα, AP' = m'α, BQ = mβ, etc.; 

  AE = AP + xPQ = AP' + x'P'Q'

  mα + x{(1 - m)α + mβ} = m'α + x'{(1 - m')α + m'β},

EX. 6. ABCD is a quadrilateral;

  AB = α, AC = β, AD = γ, AP = mα, BQ = m(β - α), etc. 

The condition PQ + RS = 0 gives

  (1 - m)α + m(β - α) + (1 - m)(γ - β) - mγ = 0, 

  (1 - 2m)(α - β + γ) = 0; 

The former solution is EX. 5; the latter gives ABCD a parallelogram.