Lines in 3D Math Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

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Find parametric equations for the line through (1,0,โˆ’2)(1, 0, -2) and (3,4,1)(3, 4, 1).

Solution

  1. 1
    Find the direction vector: โŸจ3โˆ’1,4โˆ’0,1โˆ’(โˆ’2)โŸฉ=โŸจ2,4,3โŸฉ\langle 3-1, 4-0, 1-(-2) \rangle = \langle 2, 4, 3 \rangle.
  2. 2
    Using the point (1,0,โˆ’2)(1, 0, -2): x=1+2tx = 1 + 2t, y=4ty = 4t, z=โˆ’2+3tz = -2 + 3t.
  3. 3
    The symmetric equations are xโˆ’12=y4=z+23\frac{x-1}{2} = \frac{y}{4} = \frac{z+2}{3}.

Answer

x=1+2t,y=4t,z=โˆ’2+3tx = 1 + 2t, \quad y = 4t, \quad z = -2 + 3t
When given two points, the direction vector is found by subtracting coordinates. Symmetric equations are an alternative form obtained by solving each parametric equation for tt and setting them equal. Either point can be used as the base point.

About Lines in 3D

Lines in three-dimensional space described using parametric equations x=x0+atx = x_0 + at, y=y0+bty = y_0 + bt, z=z0+ctz = z_0 + ct, or symmetric form xโˆ’x0a=yโˆ’y0b=zโˆ’z0c\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}, where (x0,y0,z0)(x_0, y_0, z_0) is a point on the line and โŸจa,b,cโŸฉ\langle a, b, c \rangle is the direction vector.

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