anonymous
  • anonymous
Find the point of intersection: \[r=\left(\begin{matrix}1 \\ 3\\ 1 \end{matrix}\right)+s\left(\begin{matrix}-2 \\ -1\\ 2 \end{matrix}\right),~~~~r=\left(\begin{matrix}0 \\ -2\\ 8 \end{matrix}\right)+t\left(\begin{matrix}1 \\ -1\\ 1 \end{matrix}\right)\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
the point of intersection occurs when s=2 and t=-3
anonymous
  • anonymous
|dw:1333278501730:dw|
hoblos
  • hoblos
1-2s = t --------(1) 3-s = -2-t -------(2) 1+2s = 8+t ------(3) adding (1)&(2) you get 2=8+2t then t=-3 1-2s=-3 s=2

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hoblos
  • hoblos
and if you replace in (2) you get 3-2=-2+3 which is true
anonymous
  • anonymous
(-3,1,5) is the answer, and I also got -3 and 2.... but how do you get the answer>
phi
  • phi
If you like, you can solve this using Gauss-Jordan starting with \[ {\bf r}= {\bf a_1} + s {\bf v_1}= {\bf a_2} + t {\bf v_2} \] or \[ s {\bf v_1}- t {\bf v_2} = {\bf a_1} - {\bf a_2} \] rename \( {\bf a_1} - {\bf a_2} = {\bf a_3} \) We can write this as a matrix with v as its columns times an unknown vector \[\left[\begin{matrix}{\bf v_1} & {\bf v_2} \end{matrix}\right]\left(\begin{matrix}s \\ -t\end{matrix}\right)={\bf a_3}\] or, to take care of the minus sign on the t coefficient \[\left[\begin{matrix}{\bf v_1} & {\bf -v_2} \end{matrix}\right]\left(\begin{matrix}s \\ t\end{matrix}\right)={\bf a_3}\] Now solve for the unknown vector using an augmented matrix \[\left[\begin{matrix}-2 & -1 & | &-1 \\ -1 & 1 &|& -5 \\ 2 & -1 &|& 7\end{matrix}\right]\]
anonymous
  • anonymous
does that give -3,1 and 5 though?
phi
  • phi
If you don't make any mistakes.
anonymous
  • anonymous
But, I don't know this method..?
phi
  • phi
In that case, it is the same as solving the system of equations -2s -1 t = -1 -1s +1t = -5 2s - 1t = 7
phi
  • phi
use elimination to solve
anonymous
  • anonymous
s=-3 and 2, but that doesn't give the point of intersection....
anonymous
  • anonymous
sorry, s=2 and t=-3
phi
  • phi
plug s into the first equation (or t into the 2nd)
phi
  • phi
to find r, the point of intersection
hoblos
  • hoblos
r=( 1-2s , 3-s , 1+2s) for s=2 r=(-3 , 1 , 5)

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