Linear Algebra Solutions 40

Linear Algebra Solutions 40 - P3 P4 r v  rr r P3  4  4 4...

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Unformatted text preview: P3 P4 r v  rr r P3  4  4 4  4 (a)  4 4 4 4 4 – 4 –– ––– P1 ––– – v v P4 v v ¨¨ ’ v ’v ¨¨ ¨ ’v ¨– –– ––– P1 ’ ––– v – ’ v (b) P2 P2 + = 1 2 1 2 x1 x3 y1 y3 x1 x2 y1 y2 + + x3 x4 y3 y4 x2 x3 y2 y3 + + x4 x1 y4 y1 x3 x4 y3 y4 + x4 x1 y4 y1 , after cancellation. Another possible configuration for the quadrilateral occurs when it is not convex, as in figure (b). The interior diagonal P2 P4 then gives two triangles P1 P2 P4 and P2 P3 P4 and we can proceed similarly as before. 2. ∆= a+x b+y c+z x+u y+v z+w u+a v+b w+c = a b c x y z x+u y+v z+w + x+u y+v z+w . u+a v+b w+c u+a v+b w+c Now a b c x+u y+v z+w u+a v+b w+c = = abc xyz uvw = a b c a b c x y z u v w + u+a v+b w+c u+a v+b w+c abc xyz abc abc uvw uvw abc uvw abc abc xyz. uvw + + Similarly x y z x+u y+v z+w u+a v+b w+c = xyz uvw abc 43 xyz =− a b c uvw = abc xyz. uvw ...
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