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Math Help - vectors part 2 ( new at all this)

  1. #1
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    vectors part 2 ( new at all this)

    1) Points P and Q have position vectors OP = 5i - 3j + 2k andOQ = 2i+ 3j, respectively. Find the vectorQP, and determine its direction cosines. Hence find the unit
    vector in the direction of
    QP.

    2) Find the direction cosines of the vector joining the point
    (-2,2,8) (i.e.,
    x = 2, y = 2, z = 8) to the point (1,0,2).

    3) Find the parametric vector and Cartesian equations of


    the line with direction cosines
    (2/7,-3/7, 6/7) which
    passes through the point
    A with position vector i + 2j -k.
    Calculate the coordinates of the point where the line
    meets the plane
    x = -1. A second line passes through
    the points
    B and C with respective coordinates (3,-2, 1)
    and (2, 1, 4). Find the Cartesian parametric equations of
    this line and show that the two lines intersect. Calculate
    the coordinates of the point of intersection.

    4) In a Cartestian coordinate system, point A is given by
    (-3, 0,-5) and point B by (3, 4, 3). Find the Direction
    Cosines of the line
    AB and write down the vector
    equation for this line.
    A second line passes through point C with coordinates
    (2, 5, 5), and is in the direction (2i+ 3j+ 6k). Show
    that these two lines intersect at a point D and write down
    the coordinates of D.
    Find a vector that is perpendicular to both
    AB and CD.
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  2. #2
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    Hello, raza88sab!

    1) Points P and Q have position vectors \overrightarrow{OP} \:= \:5i - 3j + 2k and \overrightarrow{OQ}\: =\: 2i + 3j, respectively.
    (a) Find the vector \overrightarrow{QP}.
    (b) Determine its direction cosines.
    (c) Find the unit vector in the direction of \overrightarrow{QP}.
    We have: . \begin{Bmatrix}\overrightarrow{OP} & = & \langle 5,\,\text{-}3,\,2\rangle \\ \overrightarrow{OQ} & = & \langle 2,\,3,\,0\rangle\end{Bmatrix}


    (a) Then: . \overrightarrow{QP} \:=\:\langle2,\,3,\,0\rangle - \langle5,\text{-}3,\,2\rangle \;=\;\langle5,\,\text{-}3,\,2\rangle


    (b) The direction cosines of \langle a,\,b,\,c\rangle are: . \begin{array}{ccc}\cos\alpha & = & \frac{a}{\sqrt{a^2+b^2+c^2}} \\ \cos\beta &=& \frac{b}{\sqrt{a^2+b^2+c^2}} \\ \cos\gamma &=& \frac{c}{\sqrt{a^2+b^2+c^2}} \end{array}
    We have: . |\overrightarrow{QP}| \:=\:\sqrt{(\text{-}3)^2 + 6^2 + (\text{-}2)^2} \:=\:7

    . . The direction cosines are: . \left[\text{-}\frac{3}{7},\:\frac{6}{7},\:\text{-}\frac{2}{7}\right]


    (c) The unit vector in the direction of \overrightarrow{QP} is: . \frac{\overrightarrow{QP}}{|\overrightarrow{QP}|} \;=\;\frac{\langle5,\,\text{-}3,\,2\rangle}{7}
    . . Therefore: . \left\langle\frac{5}{7},\:-\frac{3}{7},\:\frac{2}{7}\right\rangle

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  3. #3
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    Quote Originally Posted by raza88sab View Post
    ...

    3) Find the parametric vector and Cartesian equations of
    the line with direction cosines (2/7,-3/7, 6/7) which
    passes through the point A with position vector i + 2j -k.
    Calculate the coordinates of the point where the line
    meets the plane x = -1.
    A second line passes through
    the points B and C with respective coordinates (3,-2, 1) and (2,1, 4). Find the Cartesian parametric equations of
    this line and show that the two lines intersect. Calculate
    the coordinates of the point of intersection.

    ...
    Hi,

    the parametric equation of the line:

    <x,y,z>=<1, 2, -1> + r \cdot \left \langle \frac27 , -\frac37, \frac67 \right \rangle

    \boxed{\begin{array}{l}x=1+\frac27 \cdot r \\y=2-\frac37 \cdot r\\z=-1+\frac67 \cdot r\end{array}}

    You know that x = -1. Take the 1rst equation to calculate r: r = -7. Plug in this value to calculate the coordinates of the point of intersection P: P\left( -1, 5, -7\right)

    Equations of the 2nd line:

    \langle x,y,z\rangle=\langle 2,1,4 \rangle + s \cdot \left \langle \langle 2,1,4 \rangle - \langle 3, -2,1  \rangle \right \rangle = \langle 2,1,4 \rangle + s \cdot  \langle -1, 3, 3 \rangle

    and:

    \boxed{\begin{array}{l}x=2-s\\y=1 + 3s\\z=4+3s\end{array}}


    The direction vectors of the 2 lines are linearly independent that means the lines are not parallel.
    To calculate the point of intersection I use the first two equatios, calculate r and s and check if these 2 values satisfy the third equation too. If so, the two lines intersect:

    \left|\begin{array}{l}1+\frac27 \cdot r=2-s\\2-\frac37 \cdot r=1+3s\end{array}\right. . I've got s = -\frac13 and r = \frac{14}{3}. These two values satisfy the 3rd equation:

    -1+\frac67 \cdot r = 4+3s ~\implies~-1+\frac67 \cdot \frac{14}{3} = 4+3 \cdot \left(-\frac13\right)

    Therefore the point of intersection is:

    <x,y,z>=<1, 2, -1> + \left(\frac{14}{3}\right) \cdot \left \langle \frac27 , -\frac37, \frac67 \right \rangle = \left \langle \frac73, 0, 3 \right \rangle
    Last edited by earboth; October 15th 2007 at 11:13 AM.
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