91影视

The two lines ${\mathbf{r}}_1\left(t\right)=鉄�7,3-4t,2+6t鉄�$and ${\mathbf{r}}_2\left(t\right)=鉄�6-t,3+8t,9-5t鉄�$

The goal is to demonstrate that lines intersect and to determine the equation of a plane containing two lines.

The direction vector of the line ${\mathbf{r}}_1\left(t\right)=鉄�7,3-4t,2+6t鉄�$is ${\mathbf{d}}_1=鉄�0,-4,6鉄�$and the direction vector of the line ${\mathbf{r}}_2\left(t\right)=鉄�6-t,3+8t,9-5t鉄�$is ${\mathbf{d}}_2=鉄�-1,8,-5鉄�$

The lines ${\mathbf{r}}_1\left(t\right)=鉄�7,3-4t,2+6t鉄�$and ${\mathbf{r}}_2\left(t\right)=鉄�6-t,3+8t,9-5t鉄�$are intersecting because the direction vectors ${\mathbf{d}}_1=鉄�0,-4,6鉄�$and ${\mathbf{d}}_2=鉄�-1,8,-5鉄�$are not the scalar multiple of each other.

The normal vector $\mathbf{N}$ is given by:

The point of intersection of the two lines ${\mathbf{r}}_1\left(t\right)=鉄�7,3-4t,2+6t鉄�$and ${\mathbf{r}}_2\left(v\right)=鉄�6-v,3+8v,9-5v)$is given by:

$$\begin{gathered} 7=6-v鈥�..\left(1\right) \\ 3-4t=3+8v鈥�..\left(2\right) \\ 2+6t=9-5v鈥�鈥�\left(3\right) \end{gathered}$$

The equation $\left(1\right)$gives

$v=-1$

The equation (2) and $v=-1$gives:

$t=2$

The values $v=-1$and $t=2$satisfies the equation $\left(3\right)$

Therefore, point of intersection is:

$$\begin{gathered} P\left(\text{intersection}\right)=(7,3-4(2),2+6(2\left)\right) \\ =(7,-5,14) \end{gathered}$$

The point of intersection of two lines ${\mathbf{r}}_1\left(t\right)=鉄�7,3-4t,2+6t鉄�$and ${\mathbf{r}}_2\left(v\right)=鉄�6-v,3+8v,9-5v鉄�$is $(7,-5,14)$

The equation of the plane is:

The equation of the plane that contains the lines ${\mathbf{r}}_1\left(t\right)=鉄�7,3-4t,2+6t鉄�$and ${\mathbf{r}}_2\left(t\right)=鉄�6-t,3+8t,9-5t鉄�$ is $14x+3y+2z-111=0$