Results 1 to 12 of 12

Math Help - Vectors Application 19

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    425

    Vectors Application 19

    The vectors a=3i-4j-k and b=2i+3j-6k are the diagonals of a parallelogram. Show that this parallelogram is a rhombus, and determine the lengths of the sides and the angles between the sides.

    So I found part A, the lengths of the sides, but I am having trouble finding part B, the angles between the sides.

    For part A, the answer I have is [5(3)^0.5]/2, which is correct.

    For part B, I think i have to use the formula
    cos(theta)=(a.b)/(|a||b|), but when I plug in the given to the formula, I do not get the answer (same as back).

    cos(theta)=(6-12+6)/(5(3)^0.5)
    theta=90degrees
    what did i do wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,792
    Thanks
    1687
    Awards
    1
    Theorem: A parallelogram is a rhombus if and only if the diagonals are perpendicular.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    ok.. so how would i find the angle between the lengths of the sides of the rhombus then?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,792
    Thanks
    1687
    Awards
    1
    Quote Originally Posted by skeske1234 View Post
    ok.. so how would i find the angle between the lengths of the sides of the rhombus then?
    Suppose that c~\&~d are the adjacent vector sides of the rhombus.
    We know that \left\| c \right\| = \left\| d \right\|. WHY?
    \left\| a \right\| =\sqrt{26}< \left\| b \right\|=7 therefore b = c + d\;\& \,a = c - d. Why is that?

    Now you can solve that for \left\| c \right\|.
    Last edited by Plato; September 1st 2009 at 02:29 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    so I have
    cos(theta)=[a.b]/[7(sqrt(26))]

    how do I get a.b without getting 0 as my answer?

    what goes on the numerator?
    Last edited by skeske1234; September 1st 2009 at 03:31 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    or am i using the wrong formula

    cos(theta)=(a.b)/(|a||b|)
    =(3,-1,-1) . (2,3,-6)
    -----------------
    7(sqrt(26))
    =0
    ----
    7(sqrt(26))
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,971
    Thanks
    1635
    Quote Originally Posted by skeske1234 View Post
    The vectors a=3i-4j-k and b=2i+3j-6k are the diagonals of a parallelogram. Show that this parallelogram is a rhombus, and determine the lengths of the sides and the angles between the sides.

    So I found part A, the lengths of the sides, but I am having trouble finding part B, the angles between the sides.

    For part A, the answer I have is [5(3)^0.5]/2, which is correct.

    For part B, I think i have to use the formula
    cos(theta)=(a.b)/(|a||b|), but when I plug in the given to the formula, I do not get the answer (same as back).

    cos(theta)=(6-12+6)/(5(3)^0.5)
    theta=90degrees
    what did i do wrong?
    Did you possibly copy the problem wrong? (3\vec{i}- 4\vec{j}- \vec{k})\cdot (2\vec{i}+ 3\vec{J}- 6\vec{k})= 3(2)-(4)(3)-(1)(-6)= 6- 12+ 6= 0 so the two vectors are clearly perpendicular. The angle between the vectors is 90 degrees (or \pi/2 radians). What is the answer in your book?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    The answers they have are 72 degrees and 108 degrees, and no i did not copy the question wrong.. I will re-type without checking above:

    The vectors a=3i-4j-k and b=2i+3j-6k are the diagonals of a parallelogram. Show that this parallelogram is a rhombus and determine the lengths of the sides and the angles between the sides.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,792
    Thanks
    1687
    Awards
    1
    Quote Originally Posted by skeske1234 View Post
    The answers they have are 72 degrees and 108 degrees, and no i did not copy the question wrong.. I will re-type without checking above:

    The vectors a=3i-4j-k and b=2i+3j-6k are the diagonals of a parallelogram. Show that this parallelogram is a rhombus and determine the lengths of the sides and the angles between the sides.
    Quote Originally Posted by Plato View Post
    Suppose that c~\&~d are the adjacent vector sides of the rhombus.
    We know that \left\| c \right\| = \left\| d \right\|. WHY?
    \left\| a \right\| =\sqrt{26}< \left\| b \right\|=7 therefore b = c + d\;\& \,a = c - d. Why is that?
    Now you can solve that for \left\| c \right\|.
    From the above you see that 2c=a+b=5i-j-7k and 2d=b-a=-i+7j-5k.

    Now that is all need to find \left\| c \right\| and the angle between c~\&~d
    Last edited by Plato; September 2nd 2009 at 06:23 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    why do you have 2c=... and 2d=... and not c=... and d=...
    why is there a 2 in front? how did you get it?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,792
    Thanks
    1687
    Awards
    1
    Quote Originally Posted by skeske1234 View Post
    why do you have 2c=... and 2d=... and not c=... and d=...
    why is there a 2 in front? how did you get it?
    I ask this question, with all due respect, if you have so much trouble with simple algebra then why are you expected to work this problem?

    c = (1/2)\left( {5i - j - 7k} \right) \;\& \,d = (1/2)\left( { - 1i+7j - 5k} \right)

    \left\| c \right\| = \left\| d \right\| = \sqrt {75/4}

    The angle between them is \arccos \left( {\frac{{c \cdot d}}{{\left\| c \right\|\left\| d \right\|}}} \right) \approx 72^0 .

    Now this is the last time I will answer any question on this thread.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,971
    Thanks
    1635
    In my first answer, I had misread the question- the vectors you are given are the diagonals and they are clearly perpendicular. But you want the angles between the sides. You can get the lengths of the sides as the hypotenuses of right triangles where half the length of the diagonals are the lengths of the legs. Once you know the lengths of the sides, you can find the angles between the sides by using the "cosine law": c^2= a^2+ b^2- 2ab cos(C) where a and b are the lengths of the sides, c is the length of one of the diagonals, and C is the angle opposite that diagonal.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Vectors Application 14
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 4th 2009, 06:12 PM
  2. Vectors Application 2
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 3rd 2009, 09:26 AM
  3. Vectors Application 3
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 31st 2009, 06:58 AM
  4. Vectors Application
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 30th 2009, 05:17 PM
  5. Vectors Application
    Posted in the Geometry Forum
    Replies: 9
    Last Post: September 23rd 2006, 08:09 AM

Search Tags


/mathhelpforum @mathhelpforum