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The equation of a plane perpendicular to a vector and passing through a given point?

  1. Mar 2, 2008 #1
    Hey guys; im studying calculus and i came across a problem for which the book does not have an answer...

    how do i find the equation of a plane perpendicular to vector (2i+3j+4k) and passing though
    the point (1,2,3)

    ^^the numbers are some i just made up and no this is not a homework q....in fact i would be glad if any of you can kindly explain to me how this problem can be solved and/or if they can give me the link to some webpage where they have stuff about this

    thanks :)
     
  2. jcsd
  3. Mar 2, 2008 #2
    The scalar product or dot product, allows you to tell if two vectors are orthogonal or perpendicular among other things. So if you can visualize that the vector [tex] (2, 3, 4)^{T} [/tex] must be orthogonal to every point in the plane then the plane through the origin with normal vector [tex] (2, 3, 4)^{T} [/tex] would be given by the equation [tex] 2x + 3y +4z = 0 [/tex]. This is evident because [tex] (2, 3, 4) \bullet ( x, y, z )^{T} = 2x + 3y +4z = 0 [/tex] implies [tex] (2, 3, 4)^{T} [/tex] is orthogonal to every vector in the plane. Now I leave it to you to figure out how to shift the plane so that it goes through the specified point.
     
    Last edited: Mar 2, 2008
  4. Mar 2, 2008 #3

    tiny-tim

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    Draw the perpendicular!

    Hi starsiege! Welcome to PF! :smile:

    Hint: All lines in that plane will be perpendicular to that vector!

    So draw the line through the origin, O, parallel to (2i+3j+4k).

    You want to find the point, F, on that line which is the foot of the perpendicular from (1,2,3), which we'll call P.

    Call F (2a,3a,4a). Then what is the condition for OF to be perpendicular to FP?
     
  5. Mar 2, 2008 #4

    HallsofIvy

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    If you are expected to be able to do a problem like this, you should already know two things:

    1) A line in 3 dimensions can be written in parametric equations, x= At+ x0, y= Bt+ y0, z= Ct+ z0, where (x0, y0, z0) is point on the line and [itex]A\vec{i}+ B\vec{j}+ C\vec{k}[/itex] is parallel to the line.

    2) A plane can be written as a single equation, A(x- x0)+ B(y- y0)+ C0(z- z0)= 0 where (x0, y0, z0) is a point on the plane and [itex]A\vec{i}+ B\vec{j}+ C\vec{k}[/itex] is perpendicular to the plane.

    You are given [itex]A\vec{i}+ B\vec{j}+ C\vec{k}[/itex] and (x0, y0, z0).

    (My mistake, you don't really need to know (1) to do this problem!)
     
  6. Mar 2, 2008 #5
    thank you everyone for your help and suggestions.!

    i solved it....in fact im ashamed to see how easy it was... :redface:
    but even then i might not have found the answer if not for ur help! thanks again

    Ps: i really should stop doing math till 1 am :zzz: it took me tens of mins to even get my mind around problems at that time while it took less than a min in the morning.
     
  7. Mar 2, 2008 #6
    hehe ran into another prob as i was going through

    how do i go about a problem that gives me 3 points of a triangle A,B,C [ A (x1,y1), B(x2,y2),C(x3,y3) ] and asks me to find the interior angles of it?

    how should i start on this problem?(this is in the section that we use dot product)
     
  8. Mar 2, 2008 #7

    HallsofIvy

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    Have you learned that, in additon to the "add the products of the components" formula, [itex]\vec{u}\cdot\vec{v}= ||\vec{u}||||\vec{v}|| cos(\theta)[/itex], where [itex]\theta[/itex] is the angle between [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex]? That should do it easily.
     
  9. Mar 2, 2008 #8
    hey , thanks :) i figured it out soon after i posted the q but was not able to delete my q right away cos i was away from the comp. but thanks again.
     
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