1. ## Vector question

URGENT!URGENT!URGENT!URGENT!URGENT!URGENT!URGENT!

A dog has an initial position (20,3) and a velocity vector (0,-2)
A cat has an initial position (5,6) and a velocity vector (1,1)

Find the position vector of the cat relative to the dog ie DC

2. Hello, rhettfraser!

A dog has an initial position $(20,3)$ and a velocity vector $\langle0,-2\rangle.$

A cat has an initial position $(5,6)$ and a velocity vector $\langle1,1\rangle.$

Find the position vector of the cat relative to the dog, i.e. $\overrightarrow{DC}$

At time $t$, the cat's position vector is: $\vec C \:=\: \langle 5+t,\:6+t\rangle$

. . and the dog's position vector is: $\vec D \:=\:\langle 20,\:3-2t\rangle$

Therefore: . $\overrightarrow{DC} \;=\;\bigg\langle(5+t) - 20,\;(6+t) - (3-2t)\bigg\rangle \;=\;\langle t-15,\:3t+3\rangle$

3. ## More

Thankyou for helping

How would I use this to show that if d is the distance between the animals at any time (t) then

d2 = 10t2 – 12t + 234

4. Originally Posted by rhettfraser
Thankyou for helping

How would I use this to show that if d is the distance between the animals at any time (t) then

d2 = 10t2 – 12t + 234
d = magnitude of vector DC.

5. Originally Posted by rhettfraser
URGENT!URGENT!URGENT!URGENT!URGENT!URGENT!URGENT!

A dog has an initial position (20,3) and a velocity vector (0,-2)
A cat has an initial position (5,6) and a velocity vector (1,1)

Find the position vector of the cat relative to the dog ie DC
Are you sure this is not a question about pursuit curves?

RonL