91影视

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

${\mathcal{L}}_1.$

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

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.

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

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

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

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

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 $k=1.$

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

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

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