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Math Help - Vector question

  1. #1
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    Vector question

    Im hoping this is in right place?
    Ok, am very stuck with this =(.
    Find a vector perpendicular to both i + 2k - k and 3i -j + k.
    Im sure its very simple..i know it is but i just can't do it =/...the answer is i - 4j - 7k.

    I would greatly appreciate a poke in the right direction

    Thankyou!!
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  2. #2
    o_O
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    Say that \vec{c} is the vector that results from taking the cross product of \vec{u} and \vec{v}.

    You should recall that \vec{c} is perpendicular to both \vec{u} and \vec{v}.
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  3. #3
    Super Member Showcase_22's Avatar
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    You can find the cross product by:

    \begin{vmatrix}<br />
{i}&{j}&{k}\\ <br />
{1}&{2}&{-1}\\ <br />
{3}&{-1}&{1}<br />
\end{vmatrix}

    You can interchange the 2nd and 3rd rows, but you'll get the negative of this answer.
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  4. #4
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    Hello, AshleyT!

    Even if you didn't know about the cross product,
    . . we can still answer the question.

    Of course, it takes longer . . .

    Find a vector perpendicular to both i + 2j - k and 3i -j + k
    We are given two vectors: . \begin{array}{ccc} \vec u &=& \langle1,2,\text{-}1\rangle \\ \vec v &=& \langle 3,\text{-}1,1\rangle \end{array}

    We want a vector, \vec w \:=\:\langle a,b,c\rangle, perpendicular to both \vec u and \vec v.

    Recall that two vectors are perpendicular if their dot product is zero.


    We have:

    . . \vec u\perp\vec w \quad\Rightarrow\quad \vec u\cdot\vec w \:=\:0 \quad\Rightarrow\quad \langle 1,2,\text{-}1\rangle \cdot\langle a,b,c\rangle \:=\:0 \quad\Rightarrow\quad a + 2b - c \:=\:0\;\;{\color{blue}[1]}<br />

    . . \vec v\perp\vec w \quad\Rightarrow\quad \vec v\cdot\vec w \:=\:0 \quad\Rightarrow\quad \langle 3,\text{-}1,1\rangle\cdot\langle a,b,c\rangle \:=\:0 \quad\Rightarrow\quad 3a - b + c \:=\:0\;\;{\color{blue}[2]}


    Add [1] and [2]: . 4a + b \:=\:0 \quad\Rightarrow\quad b \:=\:\text{-}4a

    Substitute into [2]: . 3a + 4a + c \:=\:0 \quad\Rightarrow\quad c \:=\:\text{-}7a


    We have: . \begin{array}{ccc}a &=& a \\ b &=& \text{-}4a \\ c &=& \text{-}7a \end{array}

    On the right, replace a with a parameter t: . \begin{array}{ccc}a &=& t \\ b &=& \text{-}4t \\ c &=& \text{-}7t\end{array}

    This represents all vectors perpendicular to both \vec u and \vec v.


    Let t = 1 and we have: . \langle 1,\text{-}4,\text{-}7\rangle \:=\:i - 4j - 7k

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, AshleyT!

    Even if you didn't know about the cross product,
    . . we can still answer the question.

    Of course, it takes longer . . .

    We are given two vectors: . \begin{array}{ccc} \vec u &=& \langle1,2,\text{-}1\rangle \\ \vec v &=& \langle 3,\text{-}1,1\rangle \end{array}

    We want a vector, \vec w \:=\:\langle a,b,c\rangle, perpendicular to both \vec u and \vec v.

    Recall that two vectors are perpendicular if their dot product is zero.


    We have:

    . . \vec u\perp\vec w \quad\Rightarrow\quad \vec u\cdot\vec w \:=\:0 \quad\Rightarrow\quad \langle 1,2,\text{-}1\rangle \cdot\langle a,b,c\rangle \:=\:0 \quad\Rightarrow\quad a + 2b - c \:=\:0\;\;{\color{blue}[1]}<br />

    . . \vec v\perp\vec w \quad\Rightarrow\quad \vec v\cdot\vec w \:=\:0 \quad\Rightarrow\quad \langle 3,\text{-}1,1\rangle\cdot\langle a,b,c\rangle \:=\:0 \quad\Rightarrow\quad 3a - b + c \:=\:0\;\;{\color{blue}[2]}


    Add [1] and [2]: . 4a + b \:=\:0 \quad\Rightarrow\quad b \:=\:\text{-}4a

    Substitute into [2]: . 3a + 4a + c \:=\:0 \quad\Rightarrow\quad c \:=\:\text{-}7a


    We have: . \begin{array}{ccc}a &=& a \\ b &=& \text{-}4a \\ c &=& \text{-}7a \end{array}

    On the right, replace a with a parameter t: . \begin{array}{ccc}a &=& t \\ b &=& \text{-}4t \\ c &=& \text{-}7t\end{array}

    This represents all vectors perpendicular to both \vec u and \vec v.


    Let t = 1 and we have: . \langle 1,\text{-}4,\text{-}7\rangle \:=\:i - 4j - 7k

    Perfect thank-you! We havn't been shown the cross product yet so we have to do things the long way =(.
    Thank-you to everyone for your replies too =).
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