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Math Help - Vectors and 3D Geometry

  1. #1
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    Vectors and 3D Geometry

    Hi guys, I am new to this forum, hope I posted this thread in the correct place...
    Anyway,I got a question on vectors which I need help in:

    The vector q is inclined at angles 135, 60 and \alpha where \alpha is an obtuse angle, to the x-,y- and z-axes respectively. The line L passes through the point A(2,-2,3) and is parallel to q. The plane 'pie' passes through B(5,-3,2) and contains the y-axis.

    (i) Find a vector EQN of line L.
    (ii) Find the cartesian EQN of plane 'pie'
    (iii) Find the acute angle b/w L and 'pie'
    (iv) Find the length of projection of AB onto 'pie'

    I cant solve (ii) and (iv). By the way, the ans to (i) is r = (2,-2,3) + \lambda (-2^1/2,1,-1) if it is of any help.

    From what i understand from 'contains y-axis', I think it means the plane is parallel to y-axis and this would mean the direction vectors of x- and z-axes is 0 right? But what does it mean? I am bad with planes + axes... I think (ii) and (iv) are related thats why I can't do (iv) as well but I do know about length of projection.

    Sorry if my post is messy. Thank you!
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  2. #2
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    Quote Originally Posted by Blizzardy View Post
    Hi guys, I am new to this forum, hope I posted this thread in the correct place...
    Anyway,I got a question on vectors which I need help in:

    The vector q is inclined at angles 135, 60 and \alpha where \alpha is an obtuse angle, to the x-,y- and z-axes respectively. The line L passes through the point A(2,-2,3) and is parallel to q. The plane 'pie' passes through B(5,-3,2) and contains the y-axis.

    ...
    (ii) Find the cartesian EQN of plane 'pie'
    ...

    From what i understand from 'contains y-axis', I think it means the plane is parallel to y-axis and this would mean the direction vectors of x- and z-axes is 0 right? But what does it mean? I am bad with planes + axes... I think (ii) and (iv) are related thats why I can't do (iv) as well but I do know about length of projection.

    Sorry if my post is messy. Thank you!
    1. You know about the plane pi (I assume that you don't refer to a kind of pastry(?)) that
    • the point B(5, -3, 2) belongs to pi
    • the y-axis belong to pi, so the origin belongs to pi too


    2. Use the origin as a fixed point of pi, then the vectors
    \vec u = \langle 0,1,0 \rangle , \vec v = \langle 5,-3,2 \rangle
    span the plane pi.

    3. A parametric equation of pi could be:

    \pi:\langle x,y,z  \rangle = s \cdot \vec u  + t \cdot \vec v

    Plug in the appropriate value.
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