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Linear algebra vector and lines in r3

  1. Sep 20, 2014 #1
    Let $$A(2,-1,1)$$, $$B$$ and $$C$$ be the vertices of a triangle where $$\overrightarrow{AB}$$ is parallel to $$\vec{v}=(2,0,-1), $$$$\overrightarrow{BC}$$ is parallel to $$\vec{w}=(1,-1,1)$$ and $$\angle(BAC)=90°$$. Find the equation of the line through $$\(A\)$$ and $$\(C\)$$ in vector and parametric forms.

    Well, there is not much I can do here. I could find the equation of the plane:
    $$-x-3y-2z=-1$$ by finding the normal of the two vectors given.
     
  2. jcsd
  3. Sep 20, 2014 #2
    I only wish to have a hint.
     
  4. Sep 20, 2014 #3

    LCKurtz

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    First, I suggest you don't use the same notation for points as vectors. I would suggest angle brackets for vectors, i.e., ##\vec v = \langle 2,0,-1\rangle##. Second, are you sure you have stated the problem correctly? As it is, there is more than one solution. For example, you could just take side AB equal to your ##\vec v##, go the right distance along the ##\vec w## direction to make AC perpendicular to AB. Once you have one solution you could extend both legs of the right triangle proportionally to a larger similar triangle. Infinitely many answers.
     
  5. Sep 21, 2014 #4

    HallsofIvy

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    (On this board use "itex" and "/itex" for formulas you want to be "inline". That would not give that strange spacing!)

    You are told that line [itex]\overline{AB}[/itex] is parallel to vector <2, 0, -1>. That should tell you immediately that line [itex]\overline{AB}[/itex] can be written in parametric form [itex]x= 2t+ x_0[/itex], [itex]y= y_0[/itex], [itex]z= -t+ z_0[/itex] where [itex](x_0, y_0, z_0)[/itex] can be any point on the line. Since the point A itself is given as (2, -1, 1), the line is [itex]x= 2t+ 2[/itex], [itex]y= -1[/itex], [itex]z= -t+ 1[/itex].

    Similarly, we can write line [itex]\overline{BC}[/itex] as [itex]x= t+ x_1[/itex], [itex]y= -t+ y_1[/itex], [itex]z= t+ z_1[/itex] but we are not given a point on
    [itex]\overline{BC}[/itex].

    We do know, however, that [itex]\overline{AC}[/itex] is perpendicular to [itex]\overline{AB}[/itex] so must be in the plane through A normal to [itex]\overline{AB}[/itex]: [itex]2(x- 2)- (z- 1)= 0[/itex] as well as lying in the same plane that [itex]\overline{BC}[/itex] is in.
     
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