MATH 2050 Linear Algebra

(Section 4)
2009 Winter

Assignment 7   -   Solutions

  1. Points   A(1,1,1), B(3,7,10) and C(7,7,8)   define a triangle in real space3.
    Find the vectors   u = AB, v = AC, w = BC   and their magnitudes ||u||, ||v|| and ||w||.
    Let   M   be the point (5,7,9).   What is the relationship between the displacement vector   m = OM   and the displacement vectors   OB and OC
    What is the exact value of the angle AMB and why?

    u = [ 2 6 9 ]T
    ||u|| = 11
    v = [ 6 6 7 ]T
    ||v|| = 11
    w = [ 4 0 -2 ]T
    ||w|| = 2 sqrt(5)
    Note that each of the three coordinates of point M is exactly half way between the corresponding coordinates of points B and C.
    OM = (OB + OC) / 2

    OM = (OB + OC) / 2

    [plot of triangle] M   is therefore the midpoint of the side   BC
    and   AM   is a median of triangle   ABC
    AB = AC = 11
    Triangle   ABC   is isosceles.
    The median through   A   therefore meets the other side   BC   at right angles.

    OR

    AM = (OM - OA) = [ 4 6 8 ]T
    AM dot BC = 0
    Line segments AM and BC therefore meet at M at right angles.

    angle AMB = 90 degrees

  1. Find the vector and parametric equations of

    1. the line parallel to the vector   d = [1 2 0]T
      that passes through the origin.
      p0 = [ 0 0 0 ]T and d = [ 1 2 0 ]T
      The vector equation of any line is   p = p0 + td
      The vector and parametric equations of the line are

      [ x y z ]T = [ 0 0 0 ]T + t [ 1 2 0 ]T
       
      x = t , y = 2t , z = 0
      t = any real number

      [The equations of this line can also be written as   y = 2x ,   z = 0 .]

    2. the line that passes through the points
      P(2,3,4) and Q(3,2,7).

      line direction vector = [ 1 -1 3 ]T
      Either of the two points P or Q may be used as the displacement vector   p0.
      The vector and parametric equations of the line are

      [ x y z ]T = [ 2 3 4 ]T + t [ 1 -1 3 ]T
       
      x = 2 + t , y = 3 - t , z = 4 + 3t
      t = any real number

  1. Find the points of intersection (if any) of the pairs of lines

    1. [ 4 5 -3 ]T + s [ 1 1 -2 ]T and [ 3 2 1 ]T + t [ 1 0 -1 ]T

      Where the lines meet, each corresponding coordinate must be the same on both lines:
      4+s = 3+t and 5+s = 2 and -3-2s = 1-t
      The middle equation requires   s = –3
      Substituting into each of the other two equations we find
      t = -2, x = 1 and z = 3

      OR

      These three equations form a linear system for (s, t):
      [ 1 -1 | -1 ] [ 1 0 | -3 ] [ -2 1 | 4 ]
      [ 1 0 | -3 ] [ 0 1 | -2 ] [ 0 0 | 0 ]
      The unique solution is   s = –3   and   t = –2
      Substituting into either line (line 2 used here):
      x = 3-2 = 1 and y = 2 and z = 1+2 = 3
      The lines therefore meet at the unique point

      (x,y,z) = (1,2,3)

    2. x = y = z; and x/3 = y/2 = z-1

      First recall that in the Cartesian symmetric equations of a line,
      (x - x0) / d1 = (y - y0) / d2 = (z - z0) / d3
      the numerator contains the coordinates of a point (x0, y0, z0) on the line and the denominator contains the Cartesian components of the line direction vector   [ d1 d2 d3 ]T .
      The Cartesian parametric form for the line follows:
      x = x0 + d1 t , y = y0 + d2 t , z = z0 + d3 t
      In parametric form the lines are therefore
      L1: x = s, y = s, z = s
      L2: x = 3t, y = 2t, z = 1 + t
      Where the lines meet, each corresponding coordinate must be the same on both lines:
      s = 3t and s = 2t and s = 1+t
      The first two simultaneous equations have only the trivial solution   s = t = 0, which is inconsistent with the third equation   s = t + 1
      There is no solution.   Therefore the lines do not meet.

      OR

      The three equations form a linear system for (s, t):
      [ 1 -3 | 0 ] [ 1 -2 | 0 ] [ 1 -1 | 1 ]
      which is obviously inconsistent.   Therefore

        the lines do not meet anywhere  

      [The direction vectors are not parallel, so the lines are not parallel either.   This is a pair of skew lines.]

      AND
      Find the angle between these two lines, correct to the nearest 0.1°.

      The angle between any two lines is the angle between their direction vectors.
      The angle   theta   between any two vectors   u   and   v   may be found from the dot product:
      u dot v = |u| |v| cos theta
      A negative value for   cos theta   means that   theta   is obtuse (the two vectors are pointing in approximately opposite directions).
      The angle between two lines is usually quoted as an angle in the range   0 <= theta <= pi/2.   Therefore use
      cos theta = |u dot v| / (|u| |v|)
      The line direction vectors of the two lines are
      u = [ 1 1 1 ]T and v = [ 3 2 1 ]T
      |u| = sqrt(3) and |v| = sqrt(14)
      cos theta = 6 / sqrt(42)
      Therefore, correct to the nearest 0.1°,

      theta = 22.2 degrees

  1. Find the projection of the vector   [ 4 1 2 ]T   on the vector   d = AB   that connects the point   A (1, 2, 3)   to the point   B (5, 3, 1) .
    Hence find the distance from the point   P (5, 3, 5)   to the line through   A   and   B.

    d = AB = [ 4 1 -2 ]T
    |d| = sqrt(21)
    The projection of   v = [ 4 1 2 ]T   on   d   is
    proj_d v = (v dot d / 21) vector d
    proj_d v = (13/21) vector d

    proj_d v = (13/21) [ 4 1 -2 ]T

    AP = [ 4 1 2 ]T
    illustration of line through A and B; N is the nearest point on the line to P Point   N   is the nearest point on the line to point   P .
    AN   is the projection of vector   AP   onto the line.
    The vector from point   P   to point   N   is
    PN = [ -32 -8 -68 ]T / 21
    PN = [ -32 -8 -68 ]T / 21
    The distance from   P   to the line is just the length of vector   PN:
    |PN| = 4/21 sqrt(357)
    or, using Pythagoras’ theorem on the right angled triangle ANP:
    |PN| = sqrt(272/21)
    |PN| = sqrt(272/21)
    Therefore the distance from   P   to the line is

    |PN| = 4 sqrt(17/21) = 3.599

  1. [From the textbook, page 180, exercises 4.2, question 36]         [10 marks]
    Circle, AOB is a diameter, C is on the circumference A   and   B   are the endpoints of a diameter of a circle, centre   O.
    Prove that if   C   is any other point on the circle,
    then the chords   AC   and   BC   are perpendicular.

    [Hint:   Express   vector AC   and   vector BC   in terms of   vector u = OA   and   vector v = OC.]

    vector AC = v - u and BC = v + u
    AC dot BC = v^2 - u^2
    OA, OB and OC are all radii of the same circle
    v = u ==> AC dot BC = 0
    Neither   vector AC   nor   vector BC   are zero vectors.
    Therefore vectors   vector AC   and   vector BC   are orthogonal and chords   AC   and   BC   are perpendicular.

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      Created 2009 02 18 and most recently modified 2009 02 27 by Dr. G.H. George