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  1. #1
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    Finding a vector

    Along the line 3x+4y=-2 in the direction of decreasing y what is the vector of this line ? I don't understand how they say the vector is (4,-3)
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  2. #2
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    Quote Originally Posted by nyasha View Post
    Along the line 3x+4y=-2 in the direction of decreasing y what is the vector of this line ?
    $\displaystyle 3x + 4y = -2$

    $\displaystyle 4y = -3x - 2$

    $\displaystyle y = -\frac{3}{4}x - \frac{1}{2}$.


    So for every $\displaystyle 4$ units you have gone to the right, you have gone $\displaystyle 3$ units down.

    This would suggest that this line would have direction the same as $\displaystyle 4\mathbf{i} - 3\mathbf{j}$.

    But we don't know the length of this vector. So multiply the vector by some parameter $\displaystyle t$.


    Therefore the vector that goes in the direction of this line that also has the same length as this line can be written as

    $\displaystyle t(4\mathbf{i} - 3\mathbf{j})$.

    Now we just need a point that lies on the line so we know where to place this vector. The $\displaystyle y$ intercept will suffice.


    Therefore, the vector form of this line is

    $\displaystyle \left(0\mathbf{i} - \frac{1}{2}\mathbf{j}\right) + t(4\mathbf{i} - 3\mathbf{j})$

    or if you like

    $\displaystyle 4t\mathbf{i} - \left(\frac{1}{2} + 3t\right)\mathbf{j}$.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    $\displaystyle 3x + 4y = -2$

    $\displaystyle 4y = -3x - 2$

    $\displaystyle y = -\frac{3}{4}x - \frac{1}{2}$.


    So for every $\displaystyle 4$ units you have gone to the right, you have gone $\displaystyle 3$ units down.

    This would suggest that this line would have direction the same as $\displaystyle 4\mathbf{i} - 3\mathbf{j}$.

    But we don't know the length of this vector. So multiply the vector by some parameter $\displaystyle t$.


    Therefore the vector that goes in the direction of this line that also has the same length as this line can be written as

    $\displaystyle t(4\mathbf{i} - 3\mathbf{j})$.

    Now we just need a point that lies on the line so we know where to place this vector. The $\displaystyle y$ intercept will suffice.


    Therefore, the vector form of this line is

    $\displaystyle \left(0\mathbf{i} - \frac{1}{2}\mathbf{j}\right) + t(4\mathbf{i} - 3\mathbf{j})$

    or if you like

    $\displaystyle 4t\mathbf{i} - \left(\frac{1}{2} + 3t\right)\mathbf{j}$.

    How did u get 4i-3j ?
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  4. #4
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    Quote Originally Posted by nyasha View Post
    How did u get 4i-3j ?
    4 units to the right implies $\displaystyle 4\mathbf{i}$.

    3 units down implies $\displaystyle -3\mathbf{j}$.
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  5. #5
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    Here's another way to do this:

    The equation of the line is 3x+4y= 2. If x= 0, 4y= 2 so y= 1/2. One point on the line is (0, 1/2). If y= 0, 3x= 2 so x= 2/3. Another point on that line is (2/3, 0).

    Since 0< 1/2, going in the direction of decreasing y means going from (0, 1/2) to (2/3, 0). A vector from (0,1/2) to (2/3, 0) is $\displaystyle (2/3- 0)\vec{i}+ (0- 1/2)\vec{j}= (2/3)\vec{i}- (1/2)\vec{j}$.

    Since multiplying a vector by any number gives another vector in that direction, we can get rid of the fractions by multiplying the vector by 6: $\displaystyle 6((2/3)\vec{i}- (1/2)\vec{j}= 4\vec{i}- 3\vec{j}$.
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