Results 1 to 3 of 3

Math Help - Vectors

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    198

    Vectors

    1. AB is a diameter of a circle centered at the origin O, and P is any point on the circumference of the circle.
    Using the position vectors of A, B and P, prove (using a scalar product) that AP is perpendicular to BP.

    2. Use a vector method to prove that the diagonals of the square OABC cross at right angles.


    Iím confused, actually how to do these?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by geton View Post
    1. AB is a diameter of a circle centered at the origin O, and P is any point on the circumference of the circle.
    Using the position vectors of A, B and P, prove (using a scalar product) that AP is perpendicular to BP.

    2. Use a vector method to prove that the diagonals of the square OABC cross at right angles.


    Iím confused, actually how to do these?
    Let \alpha denote the angle \angle(BOP)

    1. The position vector of A is (-r,0)
    The position vector of B is (r, 0)
    The position vector of P is (r \cdot \cos(\alpha), r \cdot \sin(\alpha))

    2. Then

    \overrightarrow{AP}=(r \cdot \cos(\alpha) + r, r \cdot \sin(\alpha)) and

    \overrightarrow{BP}=(r \cdot \cos(\alpha) - r, r \cdot \sin(\alpha))

    3. Now calculate

    \overrightarrow{AP} \cdot \overrightarrow{BP}= (r \cdot \cos(\alpha) + r, r \cdot \sin(\alpha)) \cdot (r \cdot \cos(\alpha) - r, r \cdot \sin(\alpha)) = r^2\cos^2(\alpha)-r^2+r^2\sin^2(\alpha) = r^2-r^2 = 0

    And therefore the legs of a right triangle are perpendicular. (Thales theorem)

    To the second question:

    The cordinates of the squares vertices are O(0, 0), A(s, 0), B(s, s), C(0, s) where s means the side length.

    Calculate the vectors of the diagonals and prove that the scalar product of these two vectors is zero.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2007
    Posts
    198
    Quote Originally Posted by earboth View Post
    Let \alpha denote the angle \angle(BOP)

    1. The position vector of A is (-r,0)
    The position vector of B is (r, 0)
    The position vector of P is (r \cdot \cos(\alpha), r \cdot \sin(\alpha))

    2. Then

    \overrightarrow{AP}=(r \cdot \cos(\alpha) + r, r \cdot \sin(\alpha)) and

    \overrightarrow{BP}=(r \cdot \cos(\alpha) - r, r \cdot \sin(\alpha))

    3. Now calculate

    \overrightarrow{AP} \cdot \overrightarrow{BP}= (r \cdot \cos(\alpha) + r, r \cdot \sin(\alpha)) \cdot (r \cdot \cos(\alpha) - r, r \cdot \sin(\alpha)) = r^2\cos^2(\alpha)-r^2+r^2\sin^2(\alpha) = r^2-r^2 = 0

    And therefore the legs of a right triangle are perpendicular. (Thales theorem)

    To the second question:

    The cordinates of the squares vertices are O(0, 0), A(s, 0), B(s, s), C(0, s) where s means the side length.

    Calculate the vectors of the diagonals and prove that the scalar product of these two vectors is zero.

    Thank you so much
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 15th 2011, 05:10 PM
  2. Replies: 3
    Last Post: June 30th 2011, 08:05 PM
  3. Replies: 2
    Last Post: June 18th 2011, 10:31 AM
  4. [SOLVED] Vectors: Finding coefficients to scalars with given vectors.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 23rd 2011, 12:47 AM
  5. Replies: 4
    Last Post: May 10th 2009, 06:03 PM

Search Tags


/mathhelpforum @mathhelpforum