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Math Help - Answers to these three vector questions I am struggling to do?

  1. #1
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    Answers to these three vector questions I am struggling to do?

    1. Suppose that the coordinates of points A, B and C with respect to a Cartesian coordinate system Oxyz are (1, -2, 2), (3, 0, -1) and (-1, 1, 4), respectively. Show that the triangle ABC is isosceles (that is, has two sides of the same length).

    2. Let P and Q be the points (2, 3, -2) and (3, 2, -2), respectively. Find the vector equation of the line PQ.

    3. What is the equation of the straight line through the points (2, -2, -1) and (4, 1, 5)? Find the shortest distance from the point (2, 1, 1) from this line.
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  2. #2
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    Hello lolz

    Welcome to Math Help Forum!
    Quote Originally Posted by lolz View Post
    1. Suppose that the coordinates of points A, B and C with respect to a Cartesian coordinate system Oxyz are (1, -2, 2), (3, 0, -1) and (-1, 1, 4), respectively. Show that the triangle ABC is isosceles (that is, has two sides of the same length).

    2. Let P and Q be the points (2, 3, -2) and (3, 2, -2), respectively. Find the vector equation of the line PQ.

    3. What is the equation of the straight line through the points (2, -2, -1) and (4, 1, 5)? Find the shortest distance from the point (2, 1, 1) from this line.
    (1) The distance between (x_1, y_1, z_1) and (x_2, y_2, z_2) is
    \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}
    Use this three times to find AB, BC, AC. You'll find two of these distances are equal.

    (2) With the usual notation:
    \vec p = 2\vec i + 3\vec j -2\vec k

    \vec q = 3\vec i +2 \vec j -2\vec k
    \vec{PQ} = \vec q -\vec p
    = \vec i -\vec j
    So if \vec r is the position vector of any point on the line PQ:
    \vec r = \vec p + \lambda\vec{PQ}, where \lambda is a variable scalar.

    \Rightarrow \vec r=2\vec i + 3\vec j -2\vec k+\lambda(\vec i -\vec j)
    This, then, is the vector equation of the line PQ.

    (3) For the first part, call the two points A and B, and then use the same technique as in (2) to find the vector equation of the line AB.

    Then, if the point (2,1,1) is C, we need to find the length of the perpendicular from C to AB. If the foot of this perpendicular is R, then:

    • You have already got the position vector of R expressed in terms of a variable \lambda, since R lies on AB.
      .


    • Now use the fact that CR \perp AB when \vec{CR} . \vec{AB} = 0 (dot product). This will give you an equation for \lambda, which you can then solve to find R, and hence the distance CR.If you want to see a recent example of this technique, there's one here.


    Can you complete these now?

    Grandad
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