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Thread: Vectors

  1. #1
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    Vectors

    1. Find two unit vectors that make an angle of 60* with v = <3,4>

    2. Find the cross product a X b for a = i + (e^t)k +(e^-t)k, b = 2i + (e^t)j - (e^-t)k

    3. Use the scalar triple product to determine whether the points A(1,3,2), B(3,-1,6),C(5,2,0), and D(3,6,-4) lie on the same plane.


    1. ?

    2. I set it up as the matrices way..
    | i j k |
    | 1 e^t e^-t |
    | 2 e^t -e^-t |

    and came out with -2i - (-e^-t - 2e^-t)j + (-e^-t-2e^t)k

    3)
    | 1 3 2 |
    | 3 -1 6 |
    | 5 2 0 |
    |3 6 -4 |

    is this the way to do this problem? I end up with like
    |0|
    |-4| if i do this.

    can someone help me with these? I think i am doing 2 and 3 incorrectly and 1 i do not know how.

    thanks.
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  2. #2
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    Hello, xfyz!

    I'm still working on #1 . . . don't know why it's giving me trouble.


    2. Find the cross product a \times b for: . a \:= \:i + e^tj + e^{-t}k,\;\;b \:= \:2i + e^tj - e^{-t}k

    I set up the matrix: . \begin{vmatrix} i & j & k \\ <br />
1 & e^t & e^{\text{-}t} \\ 2 & e^t &  \text{-}e^{\text{-}t} \end{vmatrix}

    and came out with: .  -2i - \underbrace{(-e^{\text{-}t} - 2e^{\text{-}t})}_{\text{Combine!}}j + \underbrace{(-e^{\text{-}t} - 2e^t)}_{\text{Wrong!}}k

    We have: . \begin{vmatrix}i & j & k \\ 1 & e^t & e^{\text{-}t} \\ 2 & e^t & \text{-}e^{\text{-}t} \end{vmatrix}

    . . = \;i(\text{-}1-1) -j(\text{-}e^{\text{-}t}-2e^{\text{-}t}) + k(e^t - 2e^t)

    . . = \;{\color{blue}-2i + 3e^{-t}j - e^tk}




    3. Use the scalar triple product to determine whether the points:
    A(1,3,2),\;B(3,-1,6),\;C(5,2,0),\;D(3,6,-4) lie on the same plane.
    It took me a minute to remember what the scalar triple product is for.

    Given three vectors \vec u,\:\vec v,\:\vec w, the scalar triple product, u\cdot (v \times w) gives us
    the volume of the parallelepiped ("slanted box") determined by the vectors.


    Let: . \begin{array}{ccccc}\vec{u} & = & \overrightarrow{AB} & = & \langle2,\text{-}4,4\rangle \\<br />
\vec v & = & \overrightarrow{AC} & = & \langle 4,\text{-}1,\text{-}2\rangle \\<br />
\vec{w} & = & \overrightarrow{AD} & = & \langle2,3,\text{-}6\rangle \end{array}

    The three vectors have a common point, A.
    If the vectors point in three "different" directions (like the legs of a tripod),
    . . the parallelepiped will have a volume (greater than zero).
    If the vectors are in the same plane, then the "box" will have volume zero.

    Scalar triple product: . \begin{vmatrix}2 & \text{-}4 & 4 \\ 4 & \text{-}1 & \text{-}2 \\ 2 & 3 & \text{-}6 \end{vmatrix} \;=\;2(6+6) + 4(-24+4) + 4(12+2)

    . . = \;2(12) + 4(-20) + 4(14) \;=\;24 - 80 + 56 \;=\;{\color{blue}0} . . . . There!

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  3. #3
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    Quote Originally Posted by xfyz View Post
    1. Find two unit vectors that make an angle of 60* with v = <3,4>
    For vectors u=(u_1,u_2) and v=(v_1,v_2), the scalar product u.v can be calculated in two ways. It is equal to u_1v_1+u_2v_2 and it is also equal to |u||v|\cos\theta, where θ is the angle between u and v.

    If u=(x,y),\;v=(3,4), θ = 60 and u is a unit vector, then u.v = 3x+4y (from the first of those two formulas), and u.v = 5\times{\textstyle\frac12} (from the second one, because |u|=1, |v|=\sqrt{3^2+4^2}=5 and cos(60)=1/2). Also, x^2+y^2=1 because u is a unit vector.

    Thus 3x+4y = 5/2, from which y = (5-6x)/8. Substitute that value for y into the equation x^2+y^2=1 and you'll get a quadratic equation for x ...
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