Results 1 to 5 of 5

Math Help - Find angle between vectors..given 4 points..no idea what to do

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    55

    Red face Find angle between vectors..given 4 points..no idea what to do

    Hey just really need some help with this one I have no idea where to start i want to draw it out but I don't now how to draw xyz graphs

    The question is as follows:

    A(-1,2,-2), B(2,14,2), C(2,-3,4) and D(6,-3,7) are four points in space. Find the angle between AB and CD.

    Thanks heaps
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Ok

    Quote Originally Posted by smplease View Post
    Hey just really need some help with this one I have no idea where to start i want to draw it out but I don't now how to draw xyz graphs

    The question is as follows:

    A(-1,2,-2), B(2,14,2), C(2,-3,4) and D(6,-3,7) are four points in space. Find the angle between AB and CD.

    Thanks heaps
    You dont need to per say...just use the three dimensional distance equation D=\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}...and then apply the converse of the law of cosines arccos\bigg[\frac{c^2-a^2-b^2}{-2ab}=\theta\bigg]
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2008
    Posts
    55

    Talking i can get to the distance but where do i plug them in the cosine arccos rule?

    Thanks so much for the reply... I plugged the numbers on the 3-d distance rule and got AB distance = 13 and CD distance = 5... I'm just not familiar with this cosine arccos rule.

    Where do i plug the numbers?
    And since there is 3 distances ( I'm guessing... a,b,c) do I then need to work out distance BC or am I missing the point?

    Thanks again
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,804
    Thanks
    115
    Quote Originally Posted by smplease View Post
    Thanks so much for the reply... I plugged the numbers on the 3-d distance rule and got AB distance = 13 and CD distance = 5... I'm just not familiar with this cosine arccos rule.

    Where do i plug the numbers?
    And since there is 3 distances ( I'm guessing... a,b,c) do I then need to work out distance BC or am I missing the point?

    Thanks again
    1. Calculate the vector \overrightarrow{AB}:

    \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(2,14,2) - (-1,2,-2) = (3,12,4)

    and

    \overrightarrow{CD}=(6,-3,7) - (2, -3, 4) = (4, 0, 3)

    2. Calculate the absolute value (=length) of the vectors:

    |\overrightarrow{AB}| = \sqrt{3^2+12^2+4^2}=13

    |\overrightarrow{CD}| = \sqrt{4^2+3^2}=5

    3. Now apply the formula Mathstud28 has posted:

    \cos(\theta) = \frac{(3, 12, 4) \cdot (4, 0, 3)}{13 \cdot 5} = \frac{24}{65}~\implies~ \theta \approx 68.33^\circ
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2008
    Posts
    55
    Thanks a million
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: September 4th 2009, 10:19 PM
  2. Find the angle between vectors r(t) and F(t)
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 28th 2008, 02:39 PM
  3. Replies: 1
    Last Post: April 15th 2008, 09:46 PM
  4. How can I find the angle between two vectors?
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: August 3rd 2007, 09:02 AM
  5. find angle between vectors .. confusion.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: March 12th 2007, 05:05 AM

Search Tags


/mathhelpforum @mathhelpforum