3D lines summary
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Distance between two points
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Line passing through two points
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A point P(x, y, z) is on the line L if and only if the direction numbers determined by P0 and P1 are proportional
to those determined by P1 and P2. If the proportionality constant is t we see that the conditions are:![]()
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Two points formed by the equation of a line also may be written symmetrically as:
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Two lines with slopes of (a1, b1, c1) and (a2, b2, c2) are perpendicular
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Finding the distance d between 2 lines L1 and L2 that are given by the parametric equations:
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Step (1) Calculate the cross product of the direction numbers, the result is a vector perpendicular to both lines:
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PQ = [(x1 - x0), (y1 - y0), (z1 - z0)]
Step (7) The absolute value of the dot product of n with PQ will give the required distance d between the lines:
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Direction angles, direction cosines and direction numbers
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A line has two sets of direction angles according to the pointing direction of the line![]()
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3D lines - Example 1
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Question:
Find (a) the parametric equations of the line passing through the points P1(3, 1, 1) and P2(3, 0, 2).
and also find (b) a point on the line that is located at a distance of 2 units from the point (3, 1, 1).
Solution:
a) from equation (1) we obtain the parametric line equations:![]() b) distance from any point (x, y, z) to the point (3, 1, 1) is: ![]() ![]() ![]() |
3D lines - Example 2
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Question:
Find the equation of the line that passes through the point (1, 1, ⎯ 2) and is parallel to the line
that connects the points A(1, 2, 3) and B(2, 0, 4).
Solution:
The direction numbers (values of t) of the given A B line are:
The required line that passes through point (1, 1, ⎯ 2) is:
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3D lines - Example 3
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Question:
Find the distance between the lines: L1: x = 3 + 4t, y = − 2, z = 5 − 2t and the line
L2: x = 3 + s, y = 2 + 2s, z = − 1 + 7t.
L2: x = 3 + s, y = 2 + 2s, z = − 1 + 7t.
Solution:
Step (1) Cross product of the direction numbers is:
Step (2) The norm of the vector is:
Step (4) A point P on L1 where t = 0 is at: (3, ⎯ 2, 5)
Step (5) A point Q on L2 where s = 0 is at: (3, 2, ⎯ 1)
Step (6) (Q ⎯ P) = (0, 4 ⎯ 6)
Step (7) Finally the distance between the lines is:
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Step (3) The unit vector in the line direction is: | ![]() |
Step (5) A point Q on L2 where s = 0 is at: (3, 2, ⎯ 1)
Step (6) (Q ⎯ P) = (0, 4 ⎯ 6)
Step (7) Finally the distance between the lines is:
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