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Vectors help

  1. Aug 9, 2006 #1
    Hi Guys,

    I am stuck on a maths question for my first year of Engineering at University. Your help would be greatly appreciated! :smile:

    I don't know where to start

    Thanks guys o:)
  2. jcsd
  3. Aug 9, 2006 #2


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    You need to know: The equation of a plane containing [itex](x_0,y_0,z_0)[/itex], perpendicular to the vector Ai+ Bj+ Ck, is [itex]A(x-x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex]. (For any point (x,y,z) in that plane, [itex](x-x_0)i+ (y- y_0)j+ (z- z_0)k[/itex] is in the plane and so is perpendicular to [itex]Ai+ Bj+ Ck[/itex]. The dot product of the two is 0.)

    In this problem, the vector PQ is (3-1)i+ (1-(-2))j+ (2-(-4))k= 2i+ 3j+ 6k

    As far as (b) is concerned, I'll bet there is a formula for the distance between a point and a plane in this same section of your textbook. To do it without that formula, remember that the shortest distance between a point and a plane is along a line perpendicular to that plane.
    i) Write the parametric equations, for x, y, z in terms of a paratmeter t, for a line through (-1, 1, 1) in the same direction as the vector 2i+ 3j+ 6k.
    ii) Plug those equations into the equation of the plane, from (a), to get a single equation in the single variable t and solve for t.

    iii) Put that value of t into the parametric equations to get (x,y,z) coordinates of the point on the line and plane.

    iii) Calculate the distance between that point and (-1, 1, 1).
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