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Thread: calculus 3 vector problems

  1. #1
    Junior Member
    Sep 2012
    New York

    Red face calculus 3 vector problems

    Hey everyone, i had 2 questions that i can't do for any reason

    Question 1: Gandalf the Grey started in the Forest of Mirkwood at a point with coordinates (3, -1) and arrived in the Iron Hills at the point with coordinates (5, 2). If he began walking in the direction of the vector v= 5i+ 1j and changes direction only once, when he turns at a right angle, what are the coordinates of the point where he makes the turn.

    For this i got 3.1 and 1.9 but it wasn't right. Also i know that it is dot product as the u*v=0. But i tried it converting the coordinates into unit vectors and solving R(t) and R(s) where (s,t) and than plugging (s,t) back into R(t) and R(s) to get the answer. But it isn't right.

    Question 2: The distance d of a point P to the line through points A and B is the length of the component of AP that is orthogonal to AB. so the distance from P=(-5,1,-4) to the line through the points A=(2,-1,-3) and B=(1,-5,-3) is

    For this i got 53.94 but it isn't right. i calculated AP and than AB and than used orth = u - proj. but the answer i got isn't right. please can anyone help me

    Thanks alot
    Last edited by ubhutto; Jan 30th 2014 at 07:24 PM.
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  2. #2
    MHF Contributor ebaines's Avatar
    Jun 2008

    Re: calculus 3 vector problems

    For the first one you can set up three equations in three unkowns. Let L = magnitude of the vector in the 5 \hat i+ \hat j direction, and let the second vector be x \hat i +y \hat j. The dot product of the two vectors is zero:

    5x + y = 0

    And the total x-direction displacement is 2: \frac 5 {\sqrt{26}}L + x = 2, and y-displacement: \frac L {\sqrt{26}} + y = 3.

    Solve for x and y and from that determine the turning point. If you draw a sketch you'll see that the point is greater than x=5.

    For the second problem: the projection of AP onto AB is |AP|cos(theta). This is equal to the dot product  \vec {AP} \cdot \hat {AB}, where  \hat {AB} is the unit vector in the AB direction. Using Pythagoras the length of the orthogonal is then:  \sqrt{|AP|^2 - ( \vec {AP} \cdot \hat {AB})^2 }. Is this the equation you tried to use?
    Thanks from ubhutto
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