12. Lines in R3
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Given a point P0 = (x0, y0, z0) and a direction vector v1 = 〈a,b,c〉 in R3, a line L that passes through P0 and is parallel to v1 is written parametrically as a function of t:

x(t) = x0 + at,      y(t) = y0 + bt,      z(t) = z0 + ct

Using vector notation, the same line is written

x, y, z〉 = v0 + tv1 = 〈x0, y0, z0〉 + ta,b,c〉 = 〈x0 + at, y0 + bt, z0 + ct〉,

where v0 is the vector whose head is located at P0 = (x0, y0, z0).

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Example 12.1: Find the parametric equation of a line passing through P0 = (2,−1,3) and parallel to the vector v1 = 〈5,8,−4〉.

Solution: The line is represented parametrically by

x(t) = 2 + 5t,      y(t) = −1 + 8t,      z(t) = 3 − 4t,

or in vector notation as

x, y, z〉 = 〈2,−1,3〉 + t〈5,8,−4〉 = 〈2 + 5t, −1 + 8t, 3 − 4t〉.

Note that when t = 0, we obtain the vector 〈2,−1,3〉. If the foot of this vector is placed at the origin, then its head is the ordered triple P0 = (2,−1,3)

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 A line segment from a point P0 = (x0, y0, z0) to a point P1 = (x1, y1, z1) over a ≤ t ≤ b has the form 〈x, y, z〉 = 〈x0, y0, z0〉 + [(t − a)/(b − a)]〈x1 − x0, y1 − y0, z1 − z0〉, Note that 〈x1 − x0, y1 − y0, z1 − z0〉 is the direction vector of the line.

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Example 12.2: Find the parametric equation of the line segment from P0 = (4,2,−1) to P1 = (7,−3,−2).

Solution: The direction vector v1 is found by subtracting P0 from P1:

v1 = P1 − P0 = 〈7 − 4, −3 − 2, −2 − (−1)〉 = 〈3,−5,−1〉.

Thus, the line can be written

x, y, z〉 = 〈4,2,−1〉 + t〈3,−5,−1〉 = 〈4 + 3t, 2 − 5t, −1 − t〉, for 0 ≤ t ≤ 1.

The bounds are such that t = 0 gives P0 = (4,2,−1) and t = 1 gives P1 = (7,−3,−2). Since there is a direction implied by increasing t, this is called a directed line segment.

 When a line segment between two points is constructed in this manner, the bounds on t are always 0 ≤ t ≤ 1.

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Example 12.3: Find the parametric equation of the line segment connecting P0 = (4,2,−1) and P1 = (7,−3,−2) such that t = 0 gives P0 and that t = 5 gives P1.

Solution: The difference in t-values is ba = 5 − 0 = 5. Thus, the line segment is

x, y, z〉 = 〈4,2,−1〉 + (t/5)〈3,−5,−1〉 = 〈4,2,−1〉 + t〈3/5,−1,−1/5〉 = 〈4 + (3/5)t, 2 − t, −1 − (1/5)t〉, for 0 ≤ t ≤ 5.

Note that t = 0 gives P0 and that t = 5 gives P1.

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Example 12.4: Find the parametric equation of a line segment connecting P0 = (4,2,−1) and P1 = (7,−3,−2) such that t = 4 gives P0 and t = 11 gives P1.

Solution: The starting point occurs when t = 4, indicating in a horizontal shift. The difference in t-values is ba = 11 − 4 = 7. Thus, the line segment is

x, y, z〉 = 〈4,2,−1〉 + [(t − 4)/(11 − 4)]〈3,−5,−1〉 = 〈4,2,−1〉 + 〈3(t − 4)/7, −5(t − 4)/7, −1(t − 4)/7〉
= 〈4,2,−1〉 + 〈(3/7)t − 12/7, (−5/7)t + 20/7, (−1/7)t + 4/7〉
= 〈4 − 12/7, 2 + 20/7, −1 + 4/7〉 + 〈(3/7)t, (−5/7)t, (−1/7)t
= 〈16/7 + (3/7)t, 34/7 − (5/7)t, −3/7 − (1/7)t〉, for 4 ≤ t ≤ 11.

Note that t = 4 gives P0 and that t = 11 gives P1.

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When more than one line is being considered, use different parameter variables.

Example 12.5: Let L1: 〈x, y, z〉 = 〈1,2,5〉 + t〈2,4,−3〉 and L2: 〈x, y, z〉 = 〈6,1,−2〉 + s〈4,8,−6〉 be two lines defined parametrically.

1) Are lines L1 and L2 parallel?
2) Are lines L1 and L2 the same line?

Solution:

1) The direction vector for line L1 is v1 = 〈2,4,−3〉 and the direction vector for line L2 is v2 = 〈4,8,−6〉. Since v2 = 2v1 (or equivalently, v1 = (1/2)v2), the two vectors are parallel. Thus, so are the lines.

2) Choose a point from one line and show that the other line passes through it. In this example, choose point P0 = (1,2,5) from line L1. Does L2 pass through P0 = (1,2,5)? We substitute the coordinates in P0 for x, y and z, and attempt to solve for a unique value of s that would indicate line L2 passes through a point in line L1:

1 = 6 + 4s,      2 = 1 + 8s,      5 = −2 − 6s.

From the first equation, we get s = −5/4, but from the second equation, we get s = 1/8. Since s is not unique, we conclude it is impossible that L2 passes through P0 = (1,2,5). Thus, lines L1 and L2 represent two different parallel lines.

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Example 12.6: Show that the lines L1: 〈2,3,0〉 + t〈1,−2,5〉 and L2: 〈6,−5,20〉 + s〈−3,6,−15〉 are the same line.

Solution: Their direction vectors are v1 = 〈1,−2,5〉 and v2 = 〈−3,6,−15〉. Since v2 = −3v1, the two lines are parallel. Had the direction vectors not been parallel, the parametric equations cannot possibly represent the same line.

Now, choose a point from one line, and see if it is possible to find a unique value for the parameter variable in the other line. From line L1, choose the point P0 = (2,3,0) and then attempt to find a unique value for s in L2:

2 = 6 − 3s,      3 = −5 + 6s,      0 = 20 − 15s.

From the first equation, we get s = 4/3. From the second equation, we also get s = 4/3, and from the third equation, we get s = 4/3. Since we were able to find a unique value s, we conclude that line L2 passes through a point that is in line L1. Since the lines are already parallel, this would force the two lines to be coincident, that is, the same line.

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Example 12.7: Find the point of intersection of lines L1: 〈x, y, z〉 = 〈1,2,−1〉 + t〈2,−3,4〉 and L2: 〈x, y, z〉 = 〈1,8,9〉 + s〈4,−12,−2〉.

Solution: The direction vectors are v1 = 〈2,−3,4〉 and v2 = 〈4,−12,−2〉. Since they are not scalar multiples of one another, the two lines are not parallel. To see if they intersect, we set the equations for x equal to one another, and for y, and for z:

x:      1 + 2t = 1 + 4s
y:      2 − 3t = 8 − 12s
z:      −1 + 4t = 9 − 2s.

Simplifying, we have a system of two variables in three equations:

2t − 4s = 0
−3t + 12s = 6
4t + 2s = 10.

One of two things happens: either we find a solution in s and t, in which case there is an intersection point, or we do not find a solution in s and t, in which case there is no intersection point. From the first two equations, we solve a system: Thus, we have t = 2, and back-substituting, we have s = 1. Does this solve the third equation? We substitute and simplify:

4(2) + 2(1) = 8 + 2 = 10.

We get a true statement. We were able to show that when t = 2, we generate the point (5,−4,7) on line L1, and when s = 1, we generate the same point (5,−4,7) on line L2. Thus, the two lines intersect at this point.

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 Two non-parallel and non-intersecting lines in R3 are called skew lines.

Example 12.8: Show that the lines L1: 〈x, y, z〉 = 〈1,2,−1〉 + t〈2,−3,4〉 and L2: 〈x, y, z〉 = 〈1,8,5〉 + s〈4,−12,−2〉 are skew lines.

Solution: Note that these are the same lines as from the previous example, except that a small change has been made to the equation for z in line L2. From the previous example, we established that since the direction vectors are not scalar multiples of one another, then the lines are not parallel. Next, we set the equations for x equal to one another, and for y, and for z:

x:      1 + 2t = 1 + 4s
y:      2 − 3t = 8 − 12s
z:      −1 + 4t = 5 − 2s.

This simplifies to

2t − 4s = 0
−3t + 12s = 6
4t + 2s = 6.

From the previous example, solving the system formed by the first two equations gave us t = 2 and s = 1. However, when substituting these values in the third equation, we get 4(2) + 2(1) = 6, which is false. There is no solution of this system in s and t. Thus, the two lines do not intersect, and are skew.

By Scott Surgent. Please send feedback or error notification to me at scott dot surgent at gmail.