prove that i-j , 4i-3+k, 2i-4j+5k are the vertices of triangle?
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A(x1, y1, z1) =(1,-1,0) and B(x2, y2 z2) =(4, -3 ,1) &C (x3, y3, z3) = (2 , -4 ,5)
Distance between AB=√(1-4)^2+(-1+3)^2+(0-1)^2
=√9+4+1 = √14
Squaring both side
AB^2 =14 -------1 equation
DistancebetweenBC=√(4-2)^2+(-3+4)^2+(1-5)^2
=√4+1+16= √21
Squaring both side
BC^2=21-------------2equation
DistancebetweenCA=√(2-1)^2+(-4+1)^2+(5-0)^2
=√1+9+25 =√35
Squaring both sides
CA^2 = 35---------3equation
From equation 1 and 2
AB^2+ BC^2 =14+21
=35
AB^2+ BC^2=CA^2
Hence the given point are the vertex of a triangle
Distance between AB=√(1-4)^2+(-1+3)^2+(0-1)^2
=√9+4+1 = √14
Squaring both side
AB^2 =14 -------1 equation
DistancebetweenBC=√(4-2)^2+(-3+4)^2+(1-5)^2
=√4+1+16= √21
Squaring both side
BC^2=21-------------2equation
DistancebetweenCA=√(2-1)^2+(-4+1)^2+(5-0)^2
=√1+9+25 =√35
Squaring both sides
CA^2 = 35---------3equation
From equation 1 and 2
AB^2+ BC^2 =14+21
=35
AB^2+ BC^2=CA^2
Hence the given point are the vertex of a triangle
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