# Thread: IBV23 find point P the intersection of 2 vector equations of lines

1. ## IBV23 find point P the intersection of 2 vector equations of lines

$r_1 (5 \ 1)+\lambda (3 \ -2)$ and $r_2 (-2 \ 2)+t (4 \ 1)$

presume one way to do this is turn these into line equations

$y=-\frac{2}{3}x+6$

$y=\frac{1}{4}x+\frac{3}{2}$

then $x=\frac{46}{3}$ and $y=\frac{16}{3}$

2. ## Re: IBV23 find point P the intersection of 2 vector equations of lines

Are r1 and r2 constants or parameters?

3. ## Re: IBV23 find point P the intersection of 2 vector equations of lines

Hello, bigwave!

Your equations are wrong . . .

$r_1\!:\: (5, 1)+\lambda (3 ,-2)\,\text{ and }\,r_2\!:\:(-2,2)+t (4,1)$

$\text{Find }P\text{, their point of intersection.}$

$\text{We have: }\:r_1\!:\:\begin{Bmatrix}x &=& 5 + 3\lambda \\ y &=& 1 - 2\lambda \end{Bmatrix} \qquad r_2\!:\:\begin{Bmatrix}x &=& \text{-}2+4t \\ y &=& 2 +t \end{Bmatrix}$

$\text{Equate }x's\text{ and }y's: \:\begin{Bmatrix}5+3\lambda &=& \text{-}2+4t & \;\;\Rightarrow\;\; & 3\lambda - 4t &=& \text{-}7 \\ 1 - 2\lambda &=& 2 + t & \Rightarrow & 2\lambda +t &=& \text{-}1 \end{Bmatrix}$

$\text{Solve the system of equations: }\:\lambda = \text{-}1,\;t = 1$

$\text{Therefore: }\:P(2,3)$

4. ## Re: IBV23 find point P the intersection of 2 vector equations of lines

Thanks for help
and it was r=