Vector help(Planes and normals)

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SUMMARY

The discussion focuses on finding the vector equation of the common perpendicular between the lines AB and OC, defined by the position vectors OA = i, OB = i + j, and OC = i + j + 2k. The direction of the common perpendicular is determined to be 2i - k, and the shortest distance between the lines AB and OC is established as \(\frac{2}{5}\sqrt{5}\). Additionally, the task includes deriving the equation of the plane containing line AB and the common perpendicular in the form ax + by + c = d.

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Homework Statement


With O as the origin, the points A,B,C have position vectors
i,i+j,i+j+2k
respectively, Find a vector equation of the common perpendicular of the lines AB and OC.
Show that the shortest distance between the lines AB and OC is \frac{2}{5}\sqrt{5}

Find,in the form ax+by+c=d, an equation for the plane containing AB and the common perpendicular of the lines AB and OC.


Homework Equations



Vector formulae.

The Attempt at a Solution


OA=i
OB=i+j
OC=i+j+2k

AB=AO+OB=j

the direction of the common perpendicular is ABxOC = 2i-k

But to get the vector equation I need a point on the line. How do I find that?

(Also, I was never really taught how to do these sorts of vector problems, only ones at AS math level)
 

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