Welcome to our community

Be a part of something great, join today!

[SOLVED] Find the position vector of P

karush

Well-known member
Jan 31, 2012
2,716
The vector equations of two lines are given below

$r_1=\pmatrix {5 \\ 1}+2\pmatrix {3 \\ -2}$, $r_2=\pmatrix {-2 \\ 2}+t\pmatrix {4 \\ 1}$

The lines intersect at the point \(\displaystyle P\).
Find the position vector of \(\displaystyle P\)

this being in form of \(\displaystyle r=a+tb\)
where \(\displaystyle a\) is the the position vector
and b is the direction vector.
but not sure if these needs to be converted to line equations (of which not sure how to do) or just use what is given.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, karush!

Part of your problem makes little sense.


The vector equations of two lines are given below:

. . [tex]r_1\:=\:\pmatrix {5 \\ 1}+2\pmatrix{3 \\ \text{-}2} \quad r_2\:=\:\pmatrix {\text{-}2 \\ 2}+t\pmatrix {4 \\ 1}[/tex]

The lines intersect at the point [tex]P.[/tex]
Find the position vector of [tex]P.[/tex]

this being in form of [tex]r\:=\:a+tb[/tex] . ??
where [tex]a[/tex] is the the position vector
and [tex]b[/tex] is the direction vector.
P is a point, not a line!

[tex]r_1 \cap r_2:\;{5\choose1} + s{3\choose\text{-}2} \;=\;{\text{-}2\choose2} + t{4\choose1} [/tex]

. . . . . . . . .[tex]\begin{pmatrix}5 + 3s \\ 1-2s \end{pmatrix} \;=\;\begin{pmatrix}\text{-}2 + 4t \\ 2 + t\end{pmatrix}[/tex]

. . . . . . . . . [tex]\begin{Bmatrix}5 + 3s &=& \text{-}2+4t \\ 1-2s &=& 2 + t \end{Bmatrix}[/tex]

Solve the system of equations: .[tex]\begin{Bmatrix}s &-& \text{-}1 \\ t &=& 1 \end{Bmatrix}[/tex]

Therefore: .[tex]P \:=\:{2\choose3}[/tex]
 

karush

Well-known member
Jan 31, 2012
2,716
See what you mean, however that was the way it was worded from the book. Thanks for help I am new to this topic