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Homework Help: Vector proofs for triangles and some vector plane questions

  1. Oct 6, 2008 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data

    1) Show by the use of vectors that the three altitudes of a triangle pass through the same point.

    2) Show using vectos that the bisectors of the angles of a triangle pass through thr same point.

    3)Find the distance from the point (1,0,-2) to the plane 3x-2y+z+1=0

    4) Find the distance between the lines r=(-1,-2,0)+t(1,0,-1) and r=(1,0,1)+t(1,-1,2)

    5) Find the cosine of the angle between the two planes x-2y+3z-4=0 and 2x+y-z-5=0
    2. Relevant equations



    3. The attempt at a solution
    For the first two I just need some hints on how to prove them.

    3) The equation of the plane is 3x-2y+z+1=0, meaning that the normal vector,N, to the plane is (3,-2,1). [itex]|N|=\sqrt{14}[/itex].

    So the unit vector in the direction of the normal vector:

    [tex]n= \frac{3i-2j+k}{\sqrt{14}}[/tex]

    Let P=(1,0,-2).

    I need to choose any point on the given plane.
    Let Q=(1,2,0)

    The distance I need is [itex]PQcos\theta[/itex]=|PQ.n|

    So the vector PQ=Q-P=(1,2,0)-(1,0,-2)=(0,2,2)

    therefore |PQ.n|=[(0,2,2).(3,-2,1)]/sqrt(14)

    [tex]=\frac{2}{\sqrt{14}}[/tex]

    5) The normal vector,n1 for the plane x-2y+3z-4=0 is (1,-2,3).
    The normal vector,n2 for the plane 2x+y-z-5=0 is (2,1,-1)

    Thus [itex]n_1.n_2= (1,-2,3).(2,1,-1)=-3[/itex]

    [itex]|n_1|=\sqrt{14}]/itex]
    [itex]|n_2|=\sqrt{6}[/itex]

    thus

    [tex]cos \theta = \frac{-3}{\sqrt{14} \sqrt{6}}[/tex]
     
  2. jcsd
  3. Oct 6, 2008 #2

    rock.freak667

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    4)Direction of first line,u1=(1,0,-1)
    2nd line,u2=(1,-1,2)

    [tex]\vec{OA}=(1,-2,0)[/tex]

    [tex]\vec{OB}=(1,0,1)[/tex]

    [tex]\vec{AB}=(0,2,1)[/tex]


    [tex]u_1 \times u_2=-i-3j-k \Rightarrow |u_1 \times u_2|=\sqrt{11}[/tex]

    Thus the required distance is

    [tex]|\frac{(-1,-3,-1).(0,2,1)}{\sqrt{14}}|=\frac{7}{\sqrt{11}}[/tex]

    6) Find the equation of a plane such that any point on this plane is equidistant from points (1,1,1) and (3,1,5).

    Solution:
    Midpt. AB i.e. a point on the req'd plane= ( (1+3)/2, (1+1)/2, (1+5)/2 )=(2,1,3)

    The normal vector is the vector AB=-(1,1,1)+(3,1,5)=(2,0,4). Thus (1,0,2) is also a normal vector.

    Using the defintion that (r-a).n=0

    (x-2,y-1,z-3).(1,0,2)=0
    => x-2+2z-6=0
    so x+2z=8.

    Correct?
     
    Last edited: Oct 6, 2008
  4. Oct 11, 2008 #3

    rock.freak667

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    Can someone tell me if I am going correct for the 2nd one?

    I created a vector triangle, ABC such that a+b+c=0

    Drew the bisectors,AP;BR;CQ such that AP=b/2 ; BR=c/2 and CQ=a/2.

    Finding the equation of the line passing through [itex]\vec{AP}[/itex]

    [tex]l_1: r=a+ t(a+ \frac{b}{2})[/tex]

    the line through [itex]\vec{BR}[/itex]

    [tex]l_2: r= c+ u(c+ \frac{a}{2})[/tex]

    Of they meet,then

    [tex]a+ t(a+ \frac{b}{2})= c+ u(c+ \frac{a}{2})[/tex]

    If they are equal, then a=c.

    and a(1+t)=c(1+u) as well. i.e. t=u. So they intersect. Good so far?
     
  5. Oct 11, 2008 #4

    rock.freak667

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    Here is what I tried for the 1st question. I drew a triangle ABC, with H as the point of intersection of the three altitudes.

    AH.BC=0

    (AO+OH).(BO+OC)=0
    (OA.OB-OB.OH-OA.OC+OC.OH)=0 .......1

    BH.AC=0
    (BO+OH).(AO+OC)=0
    (OA.OB-OA.OH-OC.OB+OC.OH)=0.........2

    eq'n 2-1 gives

    -OB.OH-OA.OC-OA.OH-OC.OB=0

    -OB.(OH-OC)+OA.(OH-OC)=0

    -OB.CH+OA.CH=0

    CH.(OA-OB)=0

    CH.BA=0

    Meaning that CH is perpendicular to AB. Thus all three altitudes intersect at the same point,H.

    Correct??
     
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