Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 22E from Chapter 1.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 22E
Chapter:
Problem:
Find a parametric equation of the line M through p and q. [Hint: M is parallel to the vector q - p. See the figure below.]
Step-by-Step Solution
Given Information
We are given with two vectors \[ \mathbf { p } = \left[ \begin{array} { r } { - 6 } \\ { 3 } \end{array} \right] , \mathbf { q } = \left[ \begin{array} { r } { 0 } \\ { - 4 } \end{array} \right] \] We have to find a parametric equation of the line $M$ through p and q.
Step-1: The equation of Line
If the line passes through $p$ and $q$, then it must be parallel to $p-q$, so the form of equation is: \[{\bf{x}} = p + t\left( {p - q} \right)\] Here $t$ is a parameter \[ \begin{aligned} \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] & = p + t ( q - p ) \\ & = \left[ \begin{array} { c } { - 6 } \\ { 3 } \end{array} \right] + t \left[ \begin{array} { c } { 6 } \\ { - 7 } \end{array} \right] \\ & = \left[ \begin{array} { c } { - 6 + 6 t } \\ { 3 - 7 t } \end{array} \right] \end{aligned} \] So the Parametric form of Equation is
\[\begin{array}{l} {x_1} = - 6 + 6t;\\ {x_2} = 3 - 7t \end{array}\]
We are given with two vectors \[ \mathbf { p } = \left[ \begin{array} { r } { - 6 } \\ { 3 } \end{array} \right] , \mathbf { q } = \left[ \begin{array} { r } { 0 } \\ { - 4 } \end{array} \right] \] We have to find a parametric equation of the line $M$ through p and q.
Step-1: The equation of Line
If the line passes through $p$ and $q$, then it must be parallel to $p-q$, so the form of equation is: \[{\bf{x}} = p + t\left( {p - q} \right)\] Here $t$ is a parameter \[ \begin{aligned} \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] & = p + t ( q - p ) \\ & = \left[ \begin{array} { c } { - 6 } \\ { 3 } \end{array} \right] + t \left[ \begin{array} { c } { 6 } \\ { - 7 } \end{array} \right] \\ & = \left[ \begin{array} { c } { - 6 + 6 t } \\ { 3 - 7 t } \end{array} \right] \end{aligned} \] So the Parametric form of Equation is
\[\begin{array}{l} {x_1} = - 6 + 6t;\\ {x_2} = 3 - 7t \end{array}\]