We are given three points $P(-3, 1)$, $Q(3, 4)$, and $R$ in a Cartesian plane. (i) We need to find the vector $\vec{PQ}$. (ii) We are given that $\vec{PR} = \begin{pmatrix} 2 \\ -6 \end{pmatrix}$, and we need to find the coordinates of point $R$.

1. Problem Description

We are given three points $P(-3, 1)$, $Q(3, 4)$, and $R$ in a Cartesian plane.

(i) We need to find the vector $\vec{PQ}$.

(ii) We are given that $\vec{PR} = \begin{pmatrix} 2 \\ -6 \end{pmatrix}$, and we need to find the coordinates of point $R$.

2. Solution Steps

(i) To find the vector $\vec{PQ}$, we subtract the coordinates of point $P$ from the coordinates of point $Q$:

$$\vec{PQ} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} - \begin{pmatrix} -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 - (-3) \\ 4 - 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \end{pmatrix}$$

(ii) We are given that $\vec{PR} = \begin{pmatrix} 2 \\ -6 \end{pmatrix}$. Let the coordinates of $R$ be $(x, y)$. Then

$$\vec{PR} = \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} -3 \\ 1 \end{pmatrix} = \begin{pmatrix} x + 3 \\ y - 1 \end{pmatrix}$$

Since $\vec{PR} = \begin{pmatrix} 2 \\ -6 \end{pmatrix}$, we have

$$\begin{pmatrix} x + 3 \\ y - 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -6 \end{pmatrix}$$

Equating the components, we get

$$x + 3 = 2 \implies x = 2 - 3 = -1$$

$$y - 1 = -6 \implies y = -6 + 1 = -5$$

So, the coordinates of $R$ are $(-1, -5)$.

3. Final Answer

(i) $\vec{PQ} = \begin{pmatrix} 6 \\ 3 \end{pmatrix}$

(ii) $R = (-1, -5)$