Find the position vector of P

In summary, the given problem involves two lines represented by vector equations and the task is to find the position vector of the point of intersection. The equations can be converted to a system of equations and solved to find the coordinates of the point P, which is (2,3).
  • #1
karush
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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.
 
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  • #2
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]
 
  • #3
See what you mean, however that was the way it was worded from the book. Thanks for help I am new to this topic
 

FAQ: Find the position vector of P

1. What is a position vector?

A position vector is a mathematical concept used in physics and engineering to describe the location of a point in space relative to an origin point. It is typically represented by an arrow pointing from the origin to the point, and its length corresponds to the distance between the two points.

2. How is a position vector calculated?

A position vector is calculated by subtracting the coordinates of the origin point from the coordinates of the point in question. This results in a vector with components representing the distance in each dimension from the origin to the point.

3. What is the notation for a position vector?

In mathematics, a position vector is typically denoted by the letter "r" with an arrow on top, such as r. In physics and engineering, it is often written as r = (x, y, z) or r = xi + yj + zk, where x, y, and z are the components of the vector in the x, y, and z dimensions, respectively.

4. How is a position vector used in real-world applications?

Position vectors are commonly used in physics and engineering to describe the location of objects or points in space. They are also used in navigation and robotics to determine the position of a moving object relative to a fixed point. Additionally, position vectors are used in computer graphics to render 3D objects and animations.

5. Can a position vector have negative components?

Yes, a position vector can have negative components. This indicates that the point in question is located in the negative direction of the corresponding dimension from the origin point. In other words, the point is located in the opposite direction of the arrow represented by the vector.

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