write the canonical equation of the straight line 2x+3y-2z+6=0, x-3y+z+3=0
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I’m not sure what you’re asking here, because you have provided two equations in standard form, and there are three variables, which means that the best we can do is to obtain 3 equations, each containing 2 variables. To find a unique solution (x,y,z) we would need three initial equations (system). So I’ve derived 3 linear equations (that is, three lines) each defining a relationship between a pair of variables.

(a) 2x+3y-2z+6=0

(b₁) x-3y+z+3=0 (b₂) 2x-6y+2z+6=0 (b₃) -2x+6y-2z-6=0

(c₁) 3x-z+9=0 (c₂) 4x-3y+12=0 (c₃) 9y-4z=0

The second equation has been written in 3 forms b₁, b₂ and b₃, so that, when added to (a), they produce c₁, c₂ and c₃. Each of these is a straight line in canonical form.

The picture is a 3D representation of the three 2-variable equations. The blue line 9y-4z=0 is in the vertical y-z plane which partly obscures the red line 3x-z+9=0 in the horizontal x-z plane. The lines are shown as segments for clarity, but in actual fact they should be extended from both ends. x-, y- and z-intercepts are also shown. The green line 4x-3y+12=0 is in the vertical x-y plane. The dotted lines are the axes.

by Top Rated User (1.2m points)

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