By Bobylev N. A., Bulatov V.

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Bn) of V determine the 8ame frame of reference if, and only if, there exi8t8 a nonzero 8calar z 8uch that bj = zaj for i = 0, ... , n. PROOF. Consider the frame of reference (Ao, ... , An> U). , [u] = U) and let co' ... , Cn be any homogeneous vectors for A o, ... , An' respectively. Then (co, ... , cn ) is an ordered basis of V and u = XoC o + ... + XnCn where x j '# for i = 0, ... , n. If we put aj =XjCj then [aj] = Aj (i = 0, ... , n) and ° u=ao+'" +an· If (a o, ... , an), (b o,"" bn) determine the same frame of reference, then bj = zjaj (i = 0, ...

If A, B, are three non-collinear points then the configuration consisting of these three points and their three joins is called the triangle ABO. We shall say that the two triangles ABO, A' B'O' are in perspective from a point P if the seven points P, A, B, 0, A', B ' , 0 ' are distinct and if the three lines AA' , BB' , 00' are distinct and meet at P. We then refer to P as the center of perspective of the triangles. THEOREM 3. ) If ABO, A' B'O' are two coplanar triangles in perspective from a point P, and if the pairs of corresponding sides intersect in points L, M, N, then L, M, N are distinct and collinear and the line LMN is distinct from the six sides of the triangles.

But l+m+n=O and so OL, OM, ON are coplanar. This implies that L, M, N are collinear. It follows at once from the linear indepeadence of a, b, c that L, M, N are distinct and none of the points A, B, lies on the line LMN. ° The simplicity of this proof is due to the use of homogeneous vectors. But this is not the main reason for employing them. Their real significance is that their use in the proof points the way to further results not included in the original version of the theorem. For example, if AA ' , BB', 00' are parallel instead of being concurrent, then we may still take a non-zero vector p in the line through 0 parallel to AA' , BB', 24 LINEAR GEOMETRY CHAP.