(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

See first figure.

2. Relevant equations

N/A

3. The attempt at a solution

I think I gotpart adown(See second figure) but I'm confused as to how to approachpart b.

If it has to be parallel to line [tex]l[/tex] then its going to be in the same direction as [tex]l[/tex].

In other words, if [tex]\vec{u}[/tex] is the vector:

[tex]\vec{u} = <2,1,5>[/tex]

Then the plane will be in the direction of [tex]\vec{u}[/tex].(Sorry if that sounds weird, I can't think of any other way to say it :shy:)

What property do I need to fufill in order to have my simplified equation of the plane through the point [tex]P(1,1,1)[/tex] such that it is perpendicular to the plane [tex]\Pi[/tex]?

When I read this I automatically think about the dot product of two vectors,

[tex] \vec{P}\cdot\vec{Q} = |\vec{P}||\vec{Q}|cos(\theta)[/tex]

But since they are perpendicular,

[tex]\vec{P}\cdot\vec{Q} = 0[/tex]

Any suggestions?

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# Homework Help: Lines and Planes!

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