Back to Vector-Valued Functions of One Variable

Exercise 27

Avg difficulty 2.5

Recall that any line in space may be represented parametrically by a vector-valued function.

  1. Find a vector-valued function r⃗\vr that parameterizes the line through (−2,1,4)(-2,1,4) in the direction of the vector v⃗=⟨3,2,−5⟩\vv = \langle 3, 2, -5 \rangle.

  2. Find a vector-valued function r⃗\vr that parameterizes the line of intersection of the planes x+2y−z=4x + 2y - z = 4 and 3x+y−2z=13x + y - 2z = 1.

  3. Determine the point of intersection of the lines given by x=2+3t, y=1−2t, z=4t,\begin{gathered} x = 2 + 3t, \ y = 1 - 2t, \ z = 4t, \end{gathered} x=3+1s, y=3−2s, z=2s.\begin{gathered} x = 3 + 1s, \ y = 3-2s, \ z = 2s. \end{gathered} Then, find a vector-valued function r⃗\vr that parameterizes the line that passes through the point of intersection you just found and is perpendicular to both of the given lines.

Long, Keller & Schlicker, Active Calculus Multivariable, §10.2 Vector-Valued Functions of One Variable, Exercise 27

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Adapted from Active Calculus Multivariable © 2013–2026 Nicholas Long, Mitchel T. Keller, and Steven Schlicker. Licensed under CC BY-SA 4.0. Source: Active Calculus Multivariable. Imported at revision 692c03c.