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Math Help - Finding a vector

  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 ?
    3x + 4y = -2

    4y = -3x - 2

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


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

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

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

    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 y intercept will suffice.


    Therefore, the vector form of this line is

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

    or if you like

    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
    3x + 4y = -2

    4y = -3x - 2

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


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

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

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

    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 y intercept will suffice.


    Therefore, the vector form of this line is

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

    or if you like

    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 4\mathbf{i}.

    3 units down implies -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 (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: 6((2/3)\vec{i}- (1/2)\vec{j}= 4\vec{i}- 3\vec{j}.
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