Find the position vector of P

[karush]

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 $$P$$.
Find the position vector of $$P$$

this being in form of $$r=a+tb$$
where $$a$$ 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]

Hello, karush!

Part of your problem makes little sense.

The vector equations of two lines are given below:

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

The lines intersect at the point P.
Find the position vector of P.

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

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

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

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

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

Therefore: .P \:=\:{2\choose3}

 

[karush]

See what you mean, however that was the way it was worded from the book. Thanks for help I am new to this topic