2.5 Equations of lines and planes in space  (Page 3/19)

From the symmetric equations of the line, we know that vector $\text{v}=⟨4,2,1⟩$ is a direction vector for the line. Setting the symmetric equations of the line equal to zero, we see that point $P\left(3,\mathrm{-1},3\right)$ lies on the line. Then,

To calculate the distance, we need to find $\overrightarrow{PM}\times \text{v}\text{:}$

Therefore, the distance between the point and the line is ( [link] )

1. Line $L_1$ has direction vector ${\text{v}}_1=⟨2,1,1⟩;$ line $L_2$ has direction vector ${\text{v}}_2=⟨1,3,\mathrm{-2}⟩.$ Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To determine whether the lines intersect, we see if there is a point, $\left(x,y,z\right),$ that lies on both lines. To find this point, we use the parametric equations to create a system of equalities:
   
   $2s-1=t-3;s-1=3t+8;s-4=5-2t.$
   
   By the first equation, $t=2s+2.$ Substituting into the second equation yields
   
   $\begin{array}{ccc}\hfill s-1& =\hfill & 3\left(2s+2\right)+8\hfill \\ \hfill s-1& =\hfill & 6s+6+8\hfill \\ \hfill 5s& =\hfill & \mathrm{-15}\hfill \\ \hfill s& =\hfill & \mathrm{-3.}\hfill \end{array}$
   
   Substitution into the third equation, however, yields a contradiction:
   
   $\begin{array}{ccc}\hfill s-4& =\hfill & 5-2\left(2s+2\right)\hfill \\ \hfill s-4& =\hfill & 5-4s-4\hfill \\ \hfill 5s& =\hfill & 5\hfill \\ \hfill s& =\hfill & 1.\hfill \end{array}$
   
   There is no single point that satisfies the parametric equations for $L_1\text{and}{L}_2$ simultaneously. These lines do not intersect, so they are skew (see the following figure).
   
2. Line L ₁ has direction vector ${\text{v}}_1=⟨1,\mathrm{-1},1⟩$ and passes through the origin, $\left(0,0,0\right).$ Line $L_2$ has a different direction vector, ${\text{v}}_2=⟨2,1,1⟩,$ so these lines are not parallel or equal. Let $r$ represent the parameter for line $L_1$ and let $s$ represent the parameter for $L_2\text{:}$
   
   $\begin{array}{cccc}\begin{array}{cc}x\hfill & =r\hfill \\ y\hfill & =\text{−}r\hfill \\ z\hfill & =r\hfill \end{array}\hfill & & & \begin{array}{cc}x\hfill & =2s+3\hfill \\ y\hfill & =s\hfill \\ z\hfill & =s+2.\hfill \end{array}\hfill \end{array}$
   
   Solve the system of equations to find $r=1$ and $s=-1.$ If we need to find the point of intersection, we can substitute these parameters into the original equations to get $\left(1,\mathrm{-1},1\right)$ (see the following figure).
   
3. Lines $L_1$ and $L_2$ have equivalent direction vectors: $\text{v}=⟨6,\mathrm{-2},3⟩.$ These two lines are parallel (see the following figure).