(a) Find parametric equations for the line through (5, 1, 0) that is perpendicular to the plane...

Question:

(a) Find parametric equations for the line through (5, 1, 0) that is perpendicular to the plane {eq}2x ? y + z = 1 {/eq}.

(b) In what points does this line intersect the coordinate planes?

Parametric Equation of the Line:

As a first step to determination of the parametric equation of a line is to establish the direction vector of the line which is any vector that is parallel to the line and use any point that passes through it. Thus we replace the values of the vector and the line in the formula of the parametric equations of the line . The way to determine this point of intersection of the line with the plane is to substitute the coordinates of the parametric equation of the line in the plane equation and solve the value of the variable t that satisfies the equation. After that, we substitute the value of t that we obtained in the parametric equation of the line to obtain the coordinates of the point.

Answer and Explanation:

{eq}\eqalign{ & \left( a \right)\,\,\,{\text{The parametric equation of a line passing through the point }}P\left( {{x_0},{y_0},{z_0}} \right){\text{ and direction vector }}\, \cr & \,\,\,\,\vec v = \left\langle {{v_1},{v_2},{v_3}} \right\rangle \,{\text{ is given by:}} \cr & \,\,\,\,\,\,x\left( t \right) = {x_0} + {v_1}t,\,\,\,y\left( t \right) = {y_0} + {v_2}t,\,\,\,z\left( t \right) = {z_0} + {v_3}t \cr & {\text{The equation of a plane with normal vector }}\vec n = \left\langle {a,b,c} \right\rangle {\text{ is determined by }}\,ax + by + cz = d{\text{.}} \cr & {\text{We have that the line through the point }}\,P\left( {5,1,0} \right){\text{ and is parallel to the normal (perpendicular) }} \cr & {\text{vector of the plane }}\,2x - y + z = 1. \cr & \,\,\,\,\, \Rightarrow \vec v\parallel \vec n,\,\,\,\,\,\vec n = \left\langle {2, - 1,1} \right\rangle \,\,\,\,\,\,\,\,\,\,\left[ {{\text{normal vector to the plane}}} \right] \cr & {\text{So we can use the normal vector of the plane as the direction vector of the line:}} \cr & \,\,\,\,\, \Rightarrow \vec v = \vec n = \left\langle {2, - 1,1} \right\rangle \cr & {\text{Then}}{\text{, substituting }}P\left( {{x_0},{y_0},{z_0}} \right) = \left( {5,1,0} \right)\,{\text{ and }}\vec v = \left\langle {{v_1},{v_2},{v_3}} \right\rangle = \left\langle {2, - 1,1} \right\rangle \,{\text{ in the equation of the line:}} \cr & \,\,\,\,\,\,x\left( t \right) = 5 + 2t,\,\,\,y\left( t \right) = 1 - t,\,\,\,z\left( t \right) = 0 + t \cr & {\text{Since}}{\text{, the parametric equation of the line is:}} \cr & \,\,\,\,\,\,\,\boxed{x\left( t \right) = 5 + 2t,\,\,\,y\left( t \right) = 1 - t,\,\,\,z\left( t \right) = t} \cr & \cr & \left( b \right){\text{ The equation of the }}\,xy{\text{ - plane is given by }}z = 0.{\text{ Then substituting the coordinate }}\,z{\text{ }}\,{\text{of the }} \cr & {\text{parametric equation of the line }}\,x = 5 + 2t,\,\,y = 1 - t,\,\,z = t\,{\text{ in the equation of the given plane:}} \cr & \,\,\,\,\,\,z = 0\,\,\,\, \Rightarrow t = 0 \cr & {\text{Now}}{\text{, plug the value of }}\,t = 0\,{\text{ in the equation of the line to get the coordinates of the point of }} \cr & {\text{intersection of the line with the plane:}} \cr & \,\,\,\,x = 5 + 2 \cdot 0 = 5,\,\,y = 1 - 0 = 1,\,\,z = 0 = 0 \cr & {\text{So}}{\text{, the line intersects the }}\,xy{\text{ - plane at the point: }}\boxed{\left( {5,1,0} \right)}. \cr & {\text{The equation of the }}\,yz{\text{ - plane is given by }}\,x = 0.{\text{ Then substituting the coordinate }}\,x{\text{ }}\,{\text{of the }} \cr & {\text{equation of the line }}\,{\text{ }}\,x = 5 + 2t,\,\,y = 1 - t,\,\,z = t\,{\text{ in the equation of the given plane:}} \cr & \,\,\,\,\,\,x = 0\,\,\,\, \Rightarrow 5 + 2t = 0\,\,\, \Rightarrow t = - \frac{5}{2} \cr & {\text{Now}}{\text{, set the value of }}\,t = - 2\,{\text{ in the equation of the line to get the coordinates of the point of }} \cr & {\text{intersection of the line with the plane:}} \cr & \,\,\,\,x = 5 + 2 \cdot \left( { - \frac{5}{2}} \right) = 0,\,\,y = 1 - \left( { - \frac{5}{2}} \right) = \frac{7}{2},\,\,z = - \frac{5}{2} \cr & {\text{Therefore}}{\text{, the line intersects the }}\,yz{\text{ - plane at the point}}:{\text{ }}\boxed{\left( {0,\frac{7}{2}, - \frac{5}{2}} \right)}. \cr & {\text{The line intersect the }}\,xz{\text{ - plane (}}y{\text{ = 0) when }}1 - t = 0\,\,\,\, \Rightarrow t = 1{\text{:}} \cr & \,\,\,\,x = 5 + 2 \cdot 1 = 7,\,\,y = 1 - 1 = 0,\,\,z = 1 = \cr & {\text{Thus}}{\text{, the line intersects the }}\,xz{\text{ - plane at the point}}:{\text{ }}\boxed{\left( {7,0,1} \right)} \cr} {/eq}


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Parametric Equations in Applied Contexts

from Precalculus: High School

Chapter 24 / Lesson 6
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