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Math Help - Question related to cartesian vectors

  1. #1
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    Question related to cartesian vectors

    Find two perpendicular vectors such that one of these vectors is twice as long as the other, and their sum is the vector [6,8]. I tried the following approach.

    Let's say that u = [u1,u2] and v=[v1,v2].

    Then we would have the u1v1 = -u2v2. I also got the following equations.
    u1+v1=6
    u2+v2=8
    2v1^2 + 2v2^2 = u1^2 + u2^2. I'm having trouble combining the equation to get an equation with only one variable.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Substitute u_1=6-v_1, \ u_2=6-v_2 in the first and the last equation:

    \left\{\begin{array}{ll}v_1^2+v_2^2-6v_1-6v_2=0\\v_1^2+v_2^2+12v_1+12v_2=72\end{array}\righ  t.

    Substract the first from the second:

    v_1+v_2=4\Rightarrow v_2=4-v_1

    Substitute v2: v_1^2-4v_1-4=0

    Now solve the quadratic.
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  3. #3
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    Vectors

    Hello dalbir4444
    Quote Originally Posted by dalbir4444 View Post
    Find two perpendicular vectors such that one of these vectors is twice as long as the other, and their sum is the vector [6,8]. I tried the following approach.

    Let's say that u = [u1,u2] and v=[v1,v2].

    Then we would have the u1v1 = -u2v2. I also got the following equations.
    u1+v1=6
    u2+v2=8
    2v1^2 + 2v2^2 = u1^2 + u2^2. I'm having trouble combining the equation to get an equation with only one variable.
    u_1v_1 +u_2v_2 = 0 (1)

    Also if |\vec{u}|=2|\vec{v}|, when we square both sides we get:

    u_1^2 +u_2^2 = 4(v_1^2+v_2^2) (2)

    So:

    u_1+v_1=6 \Rightarrow u_1^2 + 2u_1v_1+v_1^2 = 36

    u_2+v_2=8 \Rightarrow u_2^2 + 2u_2v_2+v_2^2 = 64

    Add and use (1): u_1^2 + u_2^2+v_1^2+v_2^2 = 100

    \Rightarrow 5(v_1^2+v_2^2) = 100, from (2)

    \Rightarrow v_1^2 + v_2^2 = 20 (3)

    Now substitute for u_1, u_2 into (1):

    (6-v_1)v_1 + (8-v_2)v_2=0

    \Rightarrow 6v_1+8v_2-(v_1^2+v_2^2)=0

    \Rightarrow 6v_1+8v_2-20=0, from (3)

    \Rightarrow v_1=\tfrac13(10-4v_2)

    Substitute into (3):

    \tfrac19(100-80v_2+16v_2^2)+v_2^2=20

    \Rightarrow 25v_2^2 -80v_2-80=0

    \Rightarrow 5v_2^2-16v_2-16=0

    \Rightarrow (5v_2+4)(v_2-4)=0

    So one solution is v_2=4 which gives \vec{v} = -2\vec{i}+4\vec{j}, \vec{u} =8\vec{i}+4\vec{j}, and another solution is v_2=-\tfrac45, which gives \vec{v} = \tfrac{22}{5}\vec{i}-\tfrac45\vec{j}, \vec{u} =\tfrac85\vec{i}+\tfrac{44}{5}\vec{j}

    Grandad
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  4. #4
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    Shouldn't u1^2 + u2^2 = 2(v1^2 + v2^2), rather than equaling 4(v1^2 +v2^2), since one of the criteria is that one of these vectors is twice as long as the other. You final answer was that . This makes |u| four times as long as |v|.
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  5. #5
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    Magnitude of a vector

    Hello dalbir4444
    Quote Originally Posted by dalbir4444 View Post
    Shouldn't u1^2 + u2^2 = 2(v1^2 + v2^2), rather than equaling 4(v1^2 +v2^2), since one of the criteria is that one of these vectors is twice as long as the other. You final answer was that . This makes |u| four times as long as |v|.
    As I said in my original posting,

    |\vec{u}| = \sqrt{u_1^2 + u_2^2}=2\sqrt{v_1^2 + v_2^2}

    \Rightarrow u_1^2 + u_2^2=4(v_1^2 + v_2^2)

    Grandad
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