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 \(\displaystyle P\).
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]

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