Results 1 to 11 of 11

Math Help - Vector Question

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Vector Question

    Hi
    i need help on the following questions:

    1)if a=4i + 3j - k, b= 2i - 4j +3k, c = 2i +k find 1.5c - b +0.5a:
    =1.5(2i+k)-2i-4j+3k+0.5(4i+3j-k)
    =3i+1.5k-2i-4j+3k+2i+1.5j-0.5k
    =3i-2.5j-2k

    Now the answer says its 3i+(11/2)j-2k, however i don't understand how they got this.

    2)In moving a crate of electrical fittings, a force F of magnitude 84 newton is applied in the direction of b, where b=3i-2j+6k
    If the create is moved point A(-2, -1, 1) to the point B(1,1,2) measured in meters, find the work done.

    W= F . \vec{AB}

    \vec{AB} = \vec{OB} - \vec{OA}
    =i+j+2k--2i-j+k
    =3i+2j+k

    i problem i have is what is the force? is it 84 or is it b=3i-2j+6k?

    3) The angle between x and y is arcos (4/21). Find the scalar t given that:
    x=6i+3j-2k and y=-2i+tj-4k

    Lastly just a side question from vectors, how would covert 1.11m/s into minutes and seconds?

    P.S
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1424
    Quote Originally Posted by Paymemoney View Post
    1)if a=4i + 3j - k, b= 2i - 4j +3k, c = 2i +k find 1.5c - b +0.5a:
    =1.5(2i+k)-2i-4j+3k+0.5(4i+3j-k)
    =3i+1.5k-2i-4j+3k+2i+1.5j-0.5k
    =3i-2.5j-2k

    Now the answer says its 3i+(11/2)j-2k, however i don't understand how they got this.
    1. It's actually

    1.5(2\mathbf{i} + \mathbf{k}) - (2\mathbf{i}- 4\mathbf{j} + 3\mathbf{k}) + 0.5(4\mathbf{i} + 3\mathbf{j} - \mathbf{k})

     = 3\mathbf{i} + 1.5\mathbf{k} -2\mathbf{i} \color{red}+ \color{black}4\mathbf{j} -3\mathbf{k} + 2\mathbf{i} + 1.5\mathbf{j} - 0.5\mathbf{k}

     = 3\mathbf{i} + 5.5\mathbf{j} -2\mathbf{k}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1424
    Quote Originally Posted by Paymemoney View Post
    2)In moving a crate of electrical fittings, a force F of magnitude 84 newton is applied in the direction of b, where b=3i-2j+6k
    If the create is moved point A(-2, -1, 1) to the point B(1,1,2) measured in meters, find the work done.

    W= F . \vec{AB}

    \vec{AB} = \vec{OB} - \vec{OA}
    =i+j+2k--2i-j+k
    =3i+2j+k

    i problem i have is what is the force? is it 84 or is it b=3i-2j+6k?
    The force is the vector that has magnitude 84 and is in the direction of \mathbf{b} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}.


    So F = k\mathbf{b}.

    We also know that |F| = 84.


    |F| = |k\mathbf{b}| = k|\mathbf{b}|.


    |\mathbf{b}| = \sqrt{3^2 + (-2)^2 + 6^2}

     = \sqrt{9 + 4 + 36}

     = \sqrt{49}

     = 7.


    Since |F| = k|\mathbf{b}|

    84 = 7k

    k = 12.



    So that means

    F = 12\mathbf{b}

     = 12(3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k})

     = 36\mathbf{i} - 24\mathbf{j} + 72\mathbf{k}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1424
    Quote Originally Posted by Paymemoney View Post
    3) The angle between x and y is arcos (4/21). Find the scalar t given that:
    x=6i+3j-2k and y=-2i+tj-4k

    Lastly just a side question from vectors, how would covert 1.11m/s into minutes and seconds?

    P.S
    3. You should know that

    \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta}, where \theta is the angle between them.


    If \mathbf{x} = 6\mathbf{i} + 3\mathbf{j} -2\mathbf{k}

    then |\mathbf{x}| = \sqrt{6^2 + 3^2 + (-2)^2}

     = \sqrt{36 + 9+ 4}

     = \sqrt{49}

     = 7.


    If \mathbf{y} = -2\mathbf{i} + t\mathbf{j} - 4\mathbf{k}

    then |\mathbf{y}| = \sqrt{(-2)^2 + t^2 + (-4)^2}

     = \sqrt{4 + t^2 + 16}

     = \sqrt{t^2 + 20}.


    \mathbf{x}\cdot \mathbf{y} = (6\mathbf{i} + 3\mathbf{j} -2\mathbf{k})\cdot (-2\mathbf{i} + t\mathbf{j} - 4\mathbf{k})

    = -12 + 3t + 8

    = 3t - 4.


    Since \mathbf{x} \cdot \mathbf{y} = |\mathbf{x}||\mathbf{y}|\cos{\theta}

    3t - 4 = 7\sqrt{t^2 + 20}\cos{\left(\arccos{\frac{4}{21}}\right)}

    3t - 4 = 7\sqrt{t^2 + 20}\cdot \frac{4}{21}

    3t - 4 = \frac{4}{3}\sqrt{t^2 + 20}

    \frac{3}{4}(3t - 4) = \sqrt{t^2 + 20}

    \frac{9}{4}t - 3 = \sqrt{t^2 + 20}

    \left(\frac{9}{4}t - 3\right)^2 = t^2 + 20

    \frac{81}{16}t^2 - \frac{27}{2}t + 9 = t^2 + 20

    \frac{65}{16}t^2 - \frac{27}{2}t - 11 = 0.


    Solve for t.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1424
    Quote Originally Posted by Paymemoney View Post
    Lastly just a side question from vectors, how would covert 1.11m/s into minutes and seconds?

    P.S
    Seeing as what you have posted is a unit of speed and not a unit of time, you can't convert it into minutes OR seconds.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2008
    Posts
    509
    Quote Originally Posted by Prove It View Post
    The force is the vector that has magnitude 84 and is in the direction of \mathbf{b} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}.


    So F = k\mathbf{b}.

    We also know that |F| = 84.


    |F| = |k\mathbf{b}| = k|\mathbf{b}|.


    |\mathbf{b}| = \sqrt{3^2 + (-2)^2 + 6^2}

     = \sqrt{9 + 4 + 36}

     = \sqrt{49}

     = 7.


    Since |F| = k|\mathbf{b}|

    84 = 7k

    k = 12.



    So that means

    F = 12\mathbf{b}

     = 12(3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k})

     = 36\mathbf{i} - 24\mathbf{j} + 72\mathbf{k}.
    i don't understand why wouldn't, 84 be the magnitude of \mathbf{b} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1424
    Quote Originally Posted by Paymemoney View Post
    i don't understand why wouldn't, 84 be the magnitude of \mathbf{b} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}
    I've shown you why. Use Pythagoras on it and work out the magnitude of \mathbf{b}. Its magnitude is 7.


    A vector has magnitude and direction. Like I said, F is a vector with a magnitude of 84 and in the DIRECTION of \mathbf{b}. F DOES NOT EQUAL \mathbf{b}. It is a SCALAR MULTIPLE of \mathbf{b}. (In other words it goes in the same direction but has been magnified to a different length). You need to figure out the magnification factor so you can multiply it by \mathbf{b} and thus FIND F.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Dec 2008
    Posts
    509
    ok, so what does the k mean in the forumla F=kb? And how did you get that forumla?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1424
    Quote Originally Posted by Paymemoney View Post
    ok, so what does the k mean in the forumla F=kb? And how did you get that forumla?
    Since F is a scalar multiple of \mathbf{b}, this means that F = k\mathbf{b}, where k is the scalar that you multiply \mathbf{b} by...
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Dec 2008
    Posts
    509
    Quote Originally Posted by Prove It View Post
    3. You should know that

    \frac{81}{16}t^2 - \frac{27}{2}t + 9 = t^2 + 20

    \frac{65}{16}t^2 - \frac{27}{2}t - 11 = 0.


    Solve for t.
    when you do these lines wouldn't you square root both sides and the result would be:

    \sqrt{\frac{81}{16}t^2-\frac{27}{2}t-11} = t
    \frac{9}{4}t^2-\sqrt{\frac{27}{2}t}-\sqrt{{11}}=0
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1424
    Quote Originally Posted by Paymemoney View Post
    when you do these lines wouldn't you square root both sides and the result would be:

    \sqrt{\frac{81}{16}t^2-\frac{27}{2}t-11} = t
    \frac{9}{4}t^2-\sqrt{\frac{27}{2}t}-\sqrt{{11}}=0
    No you definitely would not. What you end up with is a Quadratic Equation, so set it equal to 0 and use the Quadratic Formula.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Vector Question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 3rd 2011, 10:10 AM
  2. Vector Question
    Posted in the Geometry Forum
    Replies: 1
    Last Post: October 2nd 2010, 12:10 PM
  3. Vector question
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 21st 2010, 12:48 AM
  4. Need help with a vector question.
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: August 28th 2009, 05:01 AM
  5. Vector question.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 25th 2008, 10:47 AM

Search Tags


/mathhelpforum @mathhelpforum