Q .54. - Calculus

Question

. Let${\mathcal{L}}_1$and ${\mathcal{L}}_2$ be lines in${\mathrm{ℝ}}^3$, with $P_1$ and$Q_1$ points on ${\mathcal{L}}_1$ and $P_2$and $Q_2$ points on${\mathcal{L}}_2$. Show that${\mathcal{L}}_1$ is parallel to${\mathcal{L}}_2$ if and only if ${\overrightarrow{P}}_1{\overrightarrow{Q}}_1$ is parallel to $\overrightarrow{P_2{Q}_2}$.

Step-by-Step Solution

Answer

$\text{ Thus the direction vectors are equal so }\overrightarrow{P_1{Q}_1}\text{ is parallel to }{\overrightarrow{P}}_2{Q}_2\text{. }$

Step 1:Given information

$\text{ Consider }{\mathcal{L}}_1\text{ and }\mathcal{L}\text{ be lines in }{\mathrm{ℝ}}^3\text{ with }{P}_1\left(x_1,y_1,z_1\right)\text{ and }{Q}_1\left(a_1,b_1,c_1\right)\text{ be any two points on line }$

${\mathcal{L}}_1.$

$P_2\left(x_2,y_2,z_2\right)\text{ and }{Q}_2\left(a_2,b_2,c_2\right)\text{ are any two points on line }{\mathcal{L}}_2\text{. }$

Step 2:Calculation

Assume that ${\mathcal{L}}_1$ is parallel to ${\mathcal{L}}_2$.

For the lines to be parallel their direction vectors have to be parallel.

That means the direction vectors of  ${\mathcal{L}}_1$  and direction vectors of  ${\mathcal{L}}_2$ are parallel.

$\text{ Take the points }{P}_1\left(x_1,y_1,z_1\right)\text{ and }{Q}_1\left(a_1,b_1,c_1\right)\text{ of the line }{\mathcal{L}}_1\text{. }$

$\text{ The direction vectors are }{d}_1=P_1{Q}_1=\left(a_1-x_1,b_1-y_1,c_1-z_1\right)$

$\text{ Now take the points }{P}_2\left(x_2,y_2,z_2\right)\text{ and }{Q}_2\left(a_2,b_2,c_2\right)\text{ of the line }{\mathcal{L}}_2\text{. }$

$\text{ The direction vectors are }{d}_2={\overrightarrow{P}}_2{\overrightarrow{Q}}_2=\left(a_2-x_2,b_2-y_2,c_2-z_2\right)\text{. }$

When two lines are parallel then their direction vector of one equation is a scalar multiple of the direction vector of the other equation.

Thus, when the lines ${\mathcal{L}}_1$ and${\mathcal{L}}_2$are parallel then their direction vector is a scalar multiple of the other direction vector.

That is$d_1=k{d}_2$

$\left(a_1-x_1,b_1-y_1,c_1-z_1\right)=k\left(a_2-x_2,b_2-y_2,c_2-z_2\right)\text{ [since }{\mathcal{L}}_1\mathrm{isparallelto}{\mathcal{L}}_2]$

Without loss of generality assume $k=1.$

$\text{ Then, }\left(a_1-x_1,b_1-y_1,c_1-z_1\right)=\left(a_2-x_2,b_2-y_2,c_2-z_2\right)$

$\overrightarrow{P_1{Q}_1}={\overrightarrow{P}}_2{\overrightarrow{Q}}_2\left[\begin{array}{l}\text{ since }{\overrightarrow{P}}_1{\overrightarrow{Q}}_1=\left(a_1-x_1,b_1-y_1,c_1-z_1\right)\\ P_2{Q}_2=\left(a_2-x_2,b_2-y_2,c_2-z_2\right)\end{array}\right]$

Thus the direction vectors are equal so $\overrightarrow{P_1{Q}_1}$is parallel to$\overrightarrow{P_2{Q}_2}$. Hence it is proved.