Question

Let${\mathbf{r}}_1\left(t\right)={\mathbf{P}}_0+t{\mathbf{d}}_1$and ${\mathbf{r}}_2\left(u\right)={\mathbf{Q}}_0+u{\mathbf{d}}_2$respectively be the equations of lines ${\mathfrak{L}}_1$and ${\mathcal{L}}_2$Show that $\left({\mathbf{P}}_0-{\mathbf{Q}}_0\right)·\left({\mathbf{d}}_1\times {\mathbf{d}}_2\right)=0$ if and only if ${\mathcal{L}}_1$and ${\mathcal{L}}_2$lie in the same plane.

Short answer

${\mathbf{d}}_1\times {\mathbf{d}}_2$ is the normal vector to the plane containing the two lines ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$

Step-by-step solution

Given information

The equations of two lines ${\mathfrak{L}}_1$ and ${\mathcal{L}}_2$ to be ${\mathbf{r}}_1\left(t\right)={\mathbf{P}}_0+t{\mathbf{d}}_1$ and ${\mathbf{r}}_2\left(u\right)={\mathbf{Q}}_0+u{\mathbf{d}}_2$

Calculation

The objective is to show that $\left({\mathbf{P}}_0-{\mathbf{Q}}_0\right)·\left({\mathbf{d}}_1\times {\mathbf{d}}_2\right)=0$if and only if lines ${\mathcal{L}}_1$and ${\mathcal{L}}_2$lie in the same plane.

Assume that lines ${\mathcal{L}}_1$and ${\mathcal{L}}_2$ lle in the same plane. Now, prove that $\left({\mathbf{P}}_0-{\mathbf{Q}}_0\right)·\left({\mathbf{d}}_1\times {\mathbf{d}}_2\right)=0$holds.

The lines ${\mathfrak{L}}_1$ and ${\mathfrak{L}}_2$ lie in the same plane. Therefore, lines ${\mathfrak{L}}_1$ and ${\mathfrak{L}}_2$ either intersect, or are parallel or are identical.

If the lines ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$ are parallel or identical; the cross-product of direction vectors is zero.

Therefore,

${\mathbf{d}}_1\times {\mathbf{d}}_2=0$

Hence, the result $\left({\mathbf{P}}_0-{\mathbf{Q}}_0\right)·\left({\mathbf{d}}_1\times {\mathbf{d}}_2\right)=0$ holds.

Calculation

If the lines ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$ intersects, then ${\mathbf{d}}_1\times {\mathbf{d}}_2$ is orthogonal to every vector in their plane.

Therefore,

$\left({\mathbf{P}}_0-{\mathbf{Q}}_0\right)·\left({\mathbf{d}}_1\times {\mathbf{d}}_2\right)=0$

Hence, the result $\left({\mathbf{P}}_0-{\mathbf{Q}}_0\right)·\left({\mathbf{d}}_1\times {\mathbf{d}}_2\right)=0$holds.

Conversely, assume that the result $\left({\mathbf{P}}_0-{\mathbf{Q}}_0\right)·\left({\mathbf{d}}_1\times {\mathbf{d}}_2\right)=0$ holds.

The objective is to prove that the lines ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$ lie in the same plane.

The result $\left({\mathbf{P}}_0-{\mathbf{Q}}_0\right)·\left({\mathbf{d}}_1\times {\mathbf{d}}_2\right)=0$ holds. Therefore, ${\mathbf{d}}_1\times {\mathbf{d}}_2$ is orthogonal to every vector in their plane.

Therefore, ${\mathbf{d}}_1\times {\mathbf{d}}_2$ is the normal vector to the plane containing the two lines ${\mathcal{L}}_1$and ${\mathcal{L}}_2$