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Math Help - vectors: sum, parallel and perpendicular

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    vectors: sum, parallel and perpendicular

    4 vectors a, b, c and d.

    a= 3i + 2j -6k
    d= 2i - j + k

    a is the sum of vectors b and c.
    Find b and c so that b is PARALLEL to d and c is PERPENDICULAR to d.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jenny View Post
    4 vectors a, b, c and d.

    a= 3i + 2j -6k
    d= 2i - j + k

    a is the sum of vectors b and c.
    Find b and c so that b is PARALLEL to d and c is PERPENDICULAR to d.
    Let vectors b and c be
    b = xi + yj + zk
    and
    c = pi + qj + rk

    Then
    a = b + c

    3i + 2j - 6k = (x + p)i + (y + q)j + (z + r)k

    Thus
    \begin{matrix} x + p = 3 \\ y + q = 2 \\ z + r = -6 \end{matrix}

    Now, b is parallel to d, so
    d = eb

    Thus
    2i - j + k = e(xi + yj + zk)

    Thus
    \begin{matrix}ex = 2 \\ ey = -1 \\ e = z \end{matrix}

    And finally c is perpendicular to d, so
    c \cdot d = 0

    (pi + qj + rk) \cdot (2i - j + k) = 0

    2p - q + r = 0

    So we have the conditions:

    \begin{matrix} x + p = 3 \\ y + q = 2 \\ z + r = -6 \end{matrix}
    and
    \begin{matrix}ex = 2 \\ ey = -1 \\ e = z \end{matrix}
    and
    2p - q + r = 0

    We have 7 variables in 7 unknowns. It'll take a while, but it should be solvable.

    -Dan
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  3. #3
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    Quote Originally Posted by jenny View Post
    4 vectors a, b, c and d.

    a= 3i + 2j -6k
    d= 2i - j + k

    a is the sum of vectors b and c.
    Find b and c so that b is PARALLEL to d and c is PERPENDICULAR to d.
    Vector \bold{b} is the orthogonal projection of \bold{a} on \bold{d} calculated by

     \bold{b} = \frac{\bold{a}\cdot \bold{d}}{\bold{d}\cdot\bold{d}} \bold{d} = \frac{-2}{6} \bold{d} = \frac{-1}{3} \bold{d}.

    Then set  \bold{c} = \bold{a} - \bold{b}.

    Now check that \bold{a} = \bold{b} + \bold{c} and \bold{b} is parallel to \bold{d} since it is a multiple of it.

    Finally \bold{c} is perpendicular to \bold{d} since

    \begin{aligned} \bold{c}\cdot\bold{d} &= (\bold{a} - \bold{b})\cdot\bold{d}.\\ <br />
&= \bold{a}\cdot\bold{d} - \bold{b}\cdot\bold{d} \\<br />
&= \bold{a}\cdot\bold{d} - \frac{\bold{a}\cdot \bold{d}}{\bold{d}\cdot\bold{d}} \bold{d}\cdot\bold{d} \\<br />
&= \bold{a}\cdot\bold{d} - \bold{a}\cdot\bold{d} \\<br />
&= 0.<br />
\end{aligned}
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  4. #4
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    Quote Originally Posted by topsquark View Post
    why should we automatically exclude the possibility that b might have a longer length than d?
    That method does not exclude the possibility that b might be longer than d.
    Moreover, the solution is unique.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Plato View Post
    That method does not exclude the possibility that b might be longer than d.
    Thanks for pointing that out. I wasn't looking at it right. :\

    -Dan
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