Parametric line and point
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Dr.Peterson
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What forms have you learned for the equation of a plane?
If you know a form that involves the normal vector, you might consider using a cross product to obtain the normal.
pka
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We re given: [imath]\vec{d}=<-3,5,2>,\; Q:(8,-2,6)\;P:(3,4,-1)[/imath] so the line [imath]\ell(t)=Q+t\vec{d}[/imath].
You are required to show that [imath]P\notin \ell(t)[/imath] so if [imath]\vec{N}=\overrightarrow {QP}\times \vec{d}[/imath]
So that the plane [imath]\pi:\vec{N}\cdot<x-8,y+2,z-6>=0[/imath]
[imath][/imath][imath][/imath]
You are required to show that [imath]P\notin \ell(t)[/imath] so if [imath]\vec{N}=\overrightarrow {QP}\times \vec{d}[/imath]
So that the plane [imath]\pi:\vec{N}\cdot<x-8,y+2,z-6>=0[/imath]
[imath][/imath][imath][/imath]
Ok, but what am I taking a cross product of? Would it be r(t) and a line that contains the point P?
Dr.Peterson
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That's for you to decide; the goal of an exercise like this is for you to practice thinking, not to depend on others.
But you take cross products of vectors, not lines; and a key idea is that the cross product of two vectors is perpendicular to both of them. So, what vectors do you know, and is there a vector you need to find that is perpendicular to them? Think!