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mathmari
Gold Member
MHB
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Hey! 
Using vector methods show that the distance between two non parallel lines $l_1$ and $l_2$ is given by $$d=\frac{|(\overrightarrow{v}_1 - \overrightarrow{v}_2) \cdot (\overrightarrow{ a}_1 \times \overrightarrow{a}_2)|}{||\overrightarrow{a}_1 \times \overrightarrow{a}_2||}$$ where $\vec{v}_1$ and $\vec{v}_2$ are random points of $l_1$ and $l_2$ respectively, and $\vec{a}_1$ and $\vec{a}_2$ are the directions of $l_1$ and $l_2$.
HINT:
We consider the plane that contains $l_1$ and is parallel to $l_2$. Show that $\frac{\vec{a}_1 \times \vec{a}_2}{\|\vec{a}_1 \times \vec{a}_2\|}$ is unit perpendicular to that plane. Then take the projection of $\vec{v}_2-\vec{v}_1$ to that perpendicular direction.
I have done the following: We consider the plane that contains $l_1$ and is parallel to $l_2$. That means that the plane passes through the point $\overrightarrow{v}_1$ and has as parallel vector the vector $\overrightarrow{a}$.
To find the distance between the two lines, we have to find the distance between the points $\overrightarrow{v}_1$ and $\overrightarrow{v}_2$.
The vectors $\overrightarrow{a}_1$ and $\overrightarrow{a}_2$ produce the plane, so the vector $\overrightarrow{a}_1 \times \overrightarrow{a}_2$ is perpendicular to the plane.
So, the unit perpendicular vector to the plane is $\frac{\overrightarrow{a}_1 \times \overrightarrow{a}_2}{||\overrightarrow{a}_1 \times \overrightarrow{a}_2||}$.
A vector from the plane to the point $\overrightarrow{v}_2$ is $\overrightarrow{v}_2-\overrightarrow{v}_1$.
The distance that we are looking for the length of the projection of this vector onto the normal vector to the plane.
So, $$d=\frac{|(\overrightarrow{v}_1 - \overrightarrow{v}_2) \cdot (\overrightarrow{ a}_1 \times \overrightarrow{a}_2)|}{||\overrightarrow{a}_1 \times \overrightarrow{a}_2||}$$
Is this correct?? (Wondering) Could I improve something at the formulation?? (Wondering)
Using vector methods show that the distance between two non parallel lines $l_1$ and $l_2$ is given by $$d=\frac{|(\overrightarrow{v}_1 - \overrightarrow{v}_2) \cdot (\overrightarrow{ a}_1 \times \overrightarrow{a}_2)|}{||\overrightarrow{a}_1 \times \overrightarrow{a}_2||}$$ where $\vec{v}_1$ and $\vec{v}_2$ are random points of $l_1$ and $l_2$ respectively, and $\vec{a}_1$ and $\vec{a}_2$ are the directions of $l_1$ and $l_2$.
HINT:
We consider the plane that contains $l_1$ and is parallel to $l_2$. Show that $\frac{\vec{a}_1 \times \vec{a}_2}{\|\vec{a}_1 \times \vec{a}_2\|}$ is unit perpendicular to that plane. Then take the projection of $\vec{v}_2-\vec{v}_1$ to that perpendicular direction.
I have done the following: We consider the plane that contains $l_1$ and is parallel to $l_2$. That means that the plane passes through the point $\overrightarrow{v}_1$ and has as parallel vector the vector $\overrightarrow{a}$.
To find the distance between the two lines, we have to find the distance between the points $\overrightarrow{v}_1$ and $\overrightarrow{v}_2$.
The vectors $\overrightarrow{a}_1$ and $\overrightarrow{a}_2$ produce the plane, so the vector $\overrightarrow{a}_1 \times \overrightarrow{a}_2$ is perpendicular to the plane.
So, the unit perpendicular vector to the plane is $\frac{\overrightarrow{a}_1 \times \overrightarrow{a}_2}{||\overrightarrow{a}_1 \times \overrightarrow{a}_2||}$.
A vector from the plane to the point $\overrightarrow{v}_2$ is $\overrightarrow{v}_2-\overrightarrow{v}_1$.
The distance that we are looking for the length of the projection of this vector onto the normal vector to the plane.
So, $$d=\frac{|(\overrightarrow{v}_1 - \overrightarrow{v}_2) \cdot (\overrightarrow{ a}_1 \times \overrightarrow{a}_2)|}{||\overrightarrow{a}_1 \times \overrightarrow{a}_2||}$$
Is this correct?? (Wondering) Could I improve something at the formulation?? (Wondering)
