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Given a point P_{0} = (x_{0}, y_{0}, z_{0}) and a direction vector v_{1} = 〈a,b,c〉 in R^{3}, a line L that passes through P_{0} and is parallel to v_{1} is written parametrically as a function of t: Using vector notation, the same line is written where v_{0} is the vector whose head is located at P_{0} = (x_{0}, y_{0}, z_{0}). Example 12.1: Find the parametric equation of a line passing through P_{0} = (2,−1,3) and parallel to the vector v_{1} = 〈5,8,−4〉. Solution: The line is represented parametrically by or in vector notation as 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 P_{0} = (2,−1,3)
A line segment from a point P_{0} = (x_{0}, y_{0}, z_{0}) to a point P_{1} = (x_{1}, y_{1}, z_{1}) over a ≤ t ≤ b has the form
Note that 〈x_{1} − x_{0}, y_{1} − y_{0}, z_{1} − z_{0}〉 is the direction vector of the line.
Example 12.2: Find the parametric equation of the line segment from P_{0} = (4,2,−1) to P_{1} = (7,−3,−2). Solution: The direction vector v_{1} is found by subtracting P_{0} from P_{1}: Thus, the line can be written The bounds are such that t = 0 gives P_{0} = (4,2,−1) and t = 1 gives P_{1} = (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.
Example 12.3: Find the parametric equation of the line segment connecting P_{0} = (4,2,−1) and P_{1} = (7,−3,−2) such that t = 0 gives P_{0} and that t = 5 gives P_{1}. Solution: The difference in tvalues is b − a = 5 − 0 = 5. Thus, the line segment is Note that t = 0 gives P_{0} and that t = 5 gives P_{1}. Example 12.4: Find the parametric equation of a line segment connecting P_{0} = (4,2,−1) and P_{1} = (7,−3,−2) such that t = 4 gives P_{0} and t = 11 gives P_{1}. Solution: The starting point occurs when t = 4, indicating in a horizontal shift. The difference in tvalues is b − a = 11 − 4 = 7. Thus, the line segment is = 〈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 P_{0} and that t = 11 gives P_{1}. When more than one line is being considered, use different parameter variables. Example 12.5: Let L_{1}: 〈x, y, z〉 = 〈1,2,5〉 + t〈2,4,−3〉 and L_{2}: 〈x, y, z〉 = 〈6,1,−2〉 + s〈4,8,−6〉 be two lines defined parametrically.
1) Are lines L_{1} and L_{2} parallel? Solution: 1) The direction vector for line L_{1} is v_{1} = 〈2,4,−3〉 and the direction vector for line L_{2} is v_{2} = 〈4,8,−6〉. Since v_{2} = 2v_{1} (or equivalently, v_{1} = (1/2)v_{2}), 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 P_{0} = (1,2,5) from line L_{1}. Does L_{2} pass through P_{0} = (1,2,5)? We substitute the coordinates in P_{0} for x, y and z, and attempt to solve for a unique value of s that would indicate line L_{2} passes through a point in line L_{1}: 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 L_{2} passes through P_{0} = (1,2,5). Thus, lines L_{1} and L_{2} represent two different parallel lines. Example 12.6: Show that the lines L_{1}: 〈2,3,0〉 + t〈1,−2,5〉 and L_{2}: 〈6,−5,20〉 + s〈−3,6,−15〉 are the same line. Solution: Their direction vectors are v_{1} = 〈1,−2,5〉 and v_{2} = 〈−3,6,−15〉. Since v_{2} = −3v_{1}, 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 L_{1}, choose the point P_{0} = (2,3,0) and then attempt to find a unique value for s in L_{2}: 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 L_{2} passes through a point that is in line L_{1}. Since the lines are already parallel, this would force the two lines to be coincident, that is, the same line. Example 12.7: Find the point of intersection of lines L_{1}: 〈x, y, z〉 = 〈1,2,−1〉 + t〈2,−3,4〉 and L_{2}: 〈x, y, z〉 = 〈1,8,9〉 + s〈4,−12,−2〉. Solution: The direction vectors are v_{1} = 〈2,−3,4〉 and v_{2} = 〈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: y: 2 − 3t = 8 − 12s z: −1 + 4t = 9 − 2s. Simplifying, we have a system of two variables in three equations: −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 backsubstituting, we have s = 1. Does this solve the third equation? We substitute and simplify: We get a true statement. We were able to show that when t = 2, we generate the point (5,−4,7) on line L_{1}, and when s = 1, we generate the same point (5,−4,7) on line L_{2}. Thus, the two lines intersect at this point.
Two nonparallel and nonintersecting lines in R^{3} are called skew lines.
Example 12.8: Show that the lines L_{1}: 〈x, y, z〉 = 〈1,2,−1〉 + t〈2,−3,4〉 and L_{2}: 〈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 L_{2}. 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: y: 2 − 3t = 8 − 12s z: −1 + 4t = 5 − 2s. This simplifies to −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.
