Hello everyone,

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

Thank you for your help.

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

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

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

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

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

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