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Math Help - Difficult Vector Questions?

  1. #1
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    Difficult Vector Questions?

    I'm having trouble with these problems..matter of fact I don't even know where to start..please help thanks.

    1. Find the distance of the point from the line through which points in the direction of .

    2. Find the vector which makes an angle of 60 degrees with the vector and which is of the same length as , and is counterclockwise to .

    3. You start walking from a point with coordinates (1, 1) and arrived at the point with coordinates (2, 5). If you began walking in the direction of the vector and you change direction only once, when you make a turn at a right angle, what are the coordinates of the point where you make the turn?


    I figured out 2 and 3 already. I still need help on 1. Thanks
    Last edited by Kayla_N; January 22nd 2010 at 04:05 PM.
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    Dear Kayla N,

    A\equiv{(8,-4)}\mbox{ and }B\equiv{(-8,-1)}

    AB=-16i+3j

    \mid{AB}\mid=\sqrt{(16^2+3^2)}\approx{16.28}

    (-16i+3j).(-7i+4j)=\mid{(-16i+3j)}\parallel{(-7i+4j)}\mid{Cos\theta} ; \theta\mbox{ is the acute angle between AB and (-7i+4j)}

    \theta\approx{19.13^o}

    Therefore perpendicular distance to the line from the point (8,-4) = \mid{AB}\mid{Sin19.13^o}\approx{5.33}

    Hope this helps.
    Last edited by Sudharaka; January 23rd 2010 at 12:36 AM.
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    Hello Kayla_N
    Quote Originally Posted by Kayla_N View Post
    ...
    1. Find the distance of the point from the line through which points in the direction of .
    ...I still need help on 1. Thanks
    The line in the direction -7\textbf{i}+4\textbf{j} has gradient -\tfrac47, so its equation is:
    y+1=-\tfrac47(x+8)

    i.e. 7y+1=-4x-32

    i.e. 4x+7y+33=0
    Now use the formula:
    Distance of (x_1,y_1) from ax+by+c=0 =\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}
    to get:
    \frac{|32-28+33|}{\sqrt{16+49}}\approx 4.59
    if my working is correct.

    Grandad
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    Dear Grandad,

    In your calculation there is a slight error. Instead of 7y+1=-4x-32 it should be 7y+7=-4x-32. If you proceed you would get the same result that I had achieved.

    Thank you.
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  5. #5
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    Thank you, Sudharaka.
    Quote Originally Posted by Sudharaka View Post
    Dear Grandad,

    In your calculation there is a slight error. Instead of 7y+1=-4x-32 it should be 7y+7=-4x-32. If you proceed you would get the same result that I had achieved.

    Thank you.
    I was obviously in too much of a hurry when I first did this!

    My revised answer, then, is:
    The distance of (8,-4) from 4x+7y+39 = 0 is \frac{|32-28+39|}{\sqrt{4^2+7^2}}
    =\frac{43}{\sqrt{65}}

    \approx 5.33
    which agrees with your revised answer. I think we both got it wrong first time!

    Grandad
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  6. #6
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    Thanks so much for all your help.
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