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Vector problem

  1. Mar 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Given points of a triangle: [itex]A(4,1,-2),B(2,0,0),C(-2,3,-5)[/itex]. Line [itex]p[/itex] contains point [itex]B[/itex], is orthogonal to [itex]\overline{AC}[/itex], and is coplanar with [itex]ABC[/itex]. Intersection of [itex]p[/itex] and [itex]\overline{AC}[/itex] is the point [itex]B_1[/itex].
    Find vector [itex]\overrightarrow{B_1B}[/itex].

    2. Relevant equations
    -Vector projection
    - Dot product
    -Magnitude of a vector

    3. The attempt at a solution
    [tex]proj_{\overrightarrow{AC}}\overrightarrow{AB}=\overrightarrow{AB_1}=\frac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AC}|^2}\cdot \overrightarrow{AC}[/tex]
    [tex]\overrightarrow{AB}=[-2,-1,2],\overrightarrow{AC}=[-6,2,-3],|\overrightarrow{AC}|=7[/tex]
    [tex]\overrightarrow{AB}\cdot \overrightarrow{AC}=4[/tex]
    [tex]\Rightarrow proj_{\overrightarrow{AC}}\overrightarrow{AB}=\overrightarrow{AB_1}=\left[-\frac{24}{49},\frac{8}{49},-\frac{12}{49}\right][/tex]

    From [itex]\overrightarrow{AB_1}[/itex] we can find the point [itex]B_1\Rightarrow B_1=\left(\frac{172}{49},\frac{8}{49},-\frac{-110}{49}\right)[/itex] [tex]\Rightarrow \overrightarrow{B_1B}=\left[-\frac{74}{49},-\frac{57}{49},\frac{110}{49}\right][/tex]

    Is this correct?
     
    Last edited: Mar 9, 2016
  2. jcsd
  3. Mar 9, 2016 #2
    The reasoning is good, and the answer is correct if a scalar product evaluates to 0
     
  4. Mar 9, 2016 #3

    SammyS

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    Notice that once you have, ##\ \overrightarrow{AB_1}\ ## and ##\ \overrightarrow{AB}\ ##, you can get ##\ \overrightarrow{B_1 B}\ ## from ##\ \overrightarrow{B_1 B}=\overrightarrow{B_1 A}+\overrightarrow{AB}\ ##
     
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