91影视

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

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

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

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

$$\begin{gathered} \mathbf{N}={\mathbf{d}}_1脳{\mathbf{d}}_2 \\ =鉄�-1,-4,4鉄�脳鉄�-1,2,1鉄� \end{gathered}$$

$$\begin{gathered} =鉄�-12,-3,-6鉄� \\ =鉄�4,1,2鉄� \end{gathered}$$

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

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

The equation $\left(1\right)$ and $\left(2\right)$ gives

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

Therefore, point of intersection is:

The point of intersection of two lines ${\mathbf{r}}_1\left(t\right)=鉄�3-t,4-4t,-3+4t鉄�$ and ${\mathbf{r}}_2\left(v\right)=鉄�5-v,-6+2v,-2+v)$ is $(2,0,1)$