Results 1 to 2 of 2

Thread: Length and direction of vector

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    80

    Length and direction of vector

    I just want to make sure my answer looks fine, but there is there a way to break it down.

    See attachmentMath1.rtf
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    4
    Quote Originally Posted by sderosa518 View Post
    I just want to make sure my answer looks fine, but there is there a way to break it down.

    See attachmentMath1.rtf
    Dear sderosa518,

    If you have a vector $\displaystyle \underline{a}$ the direction is represented by the unit vector along that direction. That is, $\displaystyle \displaystyle\hat{n}=\frac{\underline{a}}{\mid\und erline{a}\mid}$ where $\displaystyle \mid\underline{a}\mid$ is the magnitude of the vector.

    Therefore, $\displaystyle \displaystyle\underline{a}=\mid\underline{a}\mid\f rac{\underline{a}}{\mid\underline{a}\mid}$

    In your case, $\displaystyle \underline{a}=\frac{3}{5}i-\frac{12}{5}j+\frac{4}{5}k$

    $\displaystyle \displaystyle\hat{n}=\frac{\frac{3}{5}i-\frac{12}{5}j+\frac{4}{5}k}{\sqrt{\left(\frac{3}{5 }\right)^2+\left(\frac{12}{5}\right)^2+\left(\frac {4}{5}\right)^2}}=\frac{3}{13}i-\frac{12}{13}j+\frac{4}{13}k$

    $\displaystyle \mid\underline{a}\mid=2.6$

    Therefore, $\displaystyle \underline{a}=2.6\left(\frac{3}{13}i-\frac{12}{13}j+\frac{4}{13}k\right)$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Component of vector a in direction of vector b.
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Jan 2nd 2012, 09:36 AM
  2. [SOLVED] Vectors: Find the length and direction of u * v
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Nov 15th 2010, 05:41 PM
  3. Replies: 2
    Last Post: Jan 27th 2010, 08:23 PM
  4. Replies: 4
    Last Post: Nov 5th 2009, 04:03 PM
  5. Replies: 2
    Last Post: Sep 9th 2008, 06:43 PM

Search Tags


/mathhelpforum @mathhelpforum