Intersection of common perpendicular and lines

**scherz0** · April 5th 2009, 03:16 PM

Hello everyone,

15. The common perpendicular of two skew lines with direction vectors $\vec{d_1}$ and $\vec{d_2}$ is the line that intersects both the skew lines and has direction vector $\vec{n} = \vec{d_1} + \vec{d_2}$ . Find the points of intersection of the common perpendicular with each of the lines $[x, y, z] = [0, -1, 0] + s[1, 2, 1]$ and $[d, e, f] = [-2, 2, 0] + t[2, -1, 2]$ .

Thank you for your help.

---

$\vec{n} = \vec{d_1} \times \vec{d_2}$ ,

so: $\vec{n} = [1, 2, 1] \times [2, -1, 2]$

$\vec{n} = [5, 0, -5]$

However, I am not sure how to proceed from here.

I called the common perpendicular $\vec{p}$ , whose vector equation is $\vec{p} = [a, b, c] + p[5, 0, -5]$ .

I then tried to find the points of intersection with the first given line ([x, y, z]) but I did not get useful solutions.

**Plato** · April 5th 2009, 05:15 PM

Originally Posted by scherz0

15. The common perpendicular of two skew lines with direction vectors $\vec{d_1}$ and $\vec{d_2}$ is the line that intersects both the skew lines and has direction vector $\vec{n} = \vec{d_1} + \vec{d_2}$ . Find the points of intersection of the common perpendicular with each of the lines $[x, y, z] = [0, -1, 0] + s[1, 2, 1]$ and $[d, e, f] = [-2, 2, 0] + t[2, -1, 2]$ . $\vec{n} = \vec{d_1} + \vec{d_2}$ ,

You have a very basic mistake. The direction of the common perpendicular is:
$\vec{n} = \vec{d_1} \times \vec{d_2}$

**earboth** · April 6th 2009, 05:57 AM

Originally Posted by scherz0

Hello everyone,

15. The common perpendicular of two skew lines with direction vectors $\vec{d_1}$ and $\vec{d_2}$ is the line that intersects both the skew lines and has direction vector $\vec{n} = \vec{d_1} + \vec{d_2}$ . Find the points of intersection of the common perpendicular with each of the lines $[x, y, z] = [0, -1, 0] + s[1, 2, 1]$ and $[d, e, f] = [-2, 2, 0] + t[2, -1, 2]$ .

Thank you for your help.

...

Have a look here: http://www.mathhelpforum.com/math-he...kew-lines.html

**scherz0** · April 6th 2009, 02:40 PM

Thanks for your reply.

According to to other post, I would have:

$[-2, 2, 0] - [0, -1, 0] + s \bullet [1, 2, 1] - t \bullet [2, -1, 2] = k \bullet [5, 0, -5]$

$[-2, 3, 0] + s \bullet [1, 2, 1] - t \bullet [2, -1, 2] = k \bullet [5, 0, -5]$ .

Could you please tell me how I would construct a system of equations based on this equation?

**earboth** · April 6th 2009, 10:01 PM

Originally Posted by scherz0

Thanks for your reply.

According to to other post, I would have:

$[-2, 2, 0] - [0, -1, 0] + s \bullet [1, 2, 1] - t \bullet [2, -1, 2] = k \bullet [5, 0, -5]$

$[-2, 3, 0] + s \bullet [1, 2, 1] - t \bullet [2, -1, 2] = k \bullet [5, 0, -5]$ .

Could you please tell me how I would construct a system of equations based on this equation?

$[-2, 3, 0] + s \cdot [1, 2, 1] - t \cdot [2, -1, 2] = k \cdot [5, 0, -5] ~\implies~\left\{\begin{array}{rcl}-2+s-2t&=&5k\\3+2s+t&=&0 \\ s-2t&=&-5k\end{array}\right.$

This system has the solution $(s,t,k)=\left(-1, -1, -\frac15 \right)$

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