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Math Help - vector in space

  1. #1
    Junior Member
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    Unhappy vector in space

    Question :

    describe the set of points in space whose coordinate satisfy the given equation or pair of equations:

    (i) z=2y (ii) 3x=4y, z=1

    if (i) or (ii) represents a line in space, give a unit vector that is parallel to the line. If (i) or (ii) represents a plane, give a unit vector that is normal to the plane.

    My attempt :

    z=2y is a line parallel to x-axis consisting of all points of the form (0,z,z)

    3x=4y, z=1 is a plane perpendicular to the z=1 axis consisting of all points of the form (3x,3x,1)

    (i) is a line, thus the unit vector that is parallel to the line is v/|v|= (0i-2j+k)/5

    (ii) is a plane, thus a unit vector that is normal to the plane is v/|v|= (3i-4j+k)/26

    is my answer is correct??? pls help me... tq
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  2. #2
    A Plied Mathematician
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    I'm afraid you're a bit off. A location in 3-dimensional space must be described by three variables. But a position on a line requires only one variable. A position on a plane requires two variables. So, if you're talking about a line, how many equations must you have in order to knock down the number of variables from 3 to 1? If you're talking about a plane, how many equations must you have in order to knock down the number of variables from 3 to 2?
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  3. #3
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    I am no expert as Ackbeet.

    I believe (1) is a plane because -\infty < x < \infty

    (2) is a line because z=1 and y=\frac{3}{4}x
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  4. #4
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    Hello, bobey!

    Describe the set of points in space whose coordinates
    satisfy the given equation or pair of equations:

    . . (a)\; z\:=\:2y \qquad\qquad (b)\;\begin{array}{ccc}3x&=&4y \\ z&=&1\end{array}

    if (a) or (b) represents a line in space, give a unit vector that is parallel to the line.
    If (a) or (b) represents a plane, give a unit vector that is normal to the plane.

    In (a), we have: . 2y - z \:=\:0

    This is the equation of a plane.
    It passes through the origin and has normal vector: . \vec n \:=\:\langle 0,2,-1\rangle

    The unit normal vector is:

    . . \displaystyle \frac{\vec n}{|\vec n|} \;=\;\frac{\langle 0,2,-1\rangle}{\sqrt{0^2+2^2+(\text{-}1)^2}} \;=\;\frac{\langle 0,2,-1\rangle}{\sqrt{5}} \;=\;\left\langle 0,\,\frac{2}{\sqrt{5}},\,-\frac{1}{\sqrt{5}}\right\rangle





    In (b), we have: . \begin{Bmatrix}3x &=& 4y \\ z &=& 1 \end{Bmatrix}
    We can write the equations like this: . \begin{Bmatrix} x &=& x \\ y &=& \frac{3}{4}x \\ z &=& 1 \end{Bmatrix}

    \text{On the right, replace }\,x\text{ with }\,t\!:\;\;\begin{Bmatrix}x &=& t \\ y &=& \frac{3}{4}t \\ z &=& 1\end{Bmatrix}


    These are the parametric equations of a line.

    It passes through (0,0,1) and has direction vector \vec v \:=\:\langle 1,\frac{3}{4},0\rangle

    The unit vector is:

    . . \displaystyle \frac{\vec v}{|\vec v|} \;=\;\frac{\langle 1,\frac{3}{4},0\rangle}{\sqrt{1^2+(\frac{3}{4})^2 + 0^2}} \;=\;\frac{\langle1,\frac{3}{4},0\rangle}{\frac{5}  {4}} \;=\;\left\langle \frac{4}{5},\,\frac{3}{5},\,0\right\rangle

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