The problem consists of three parts. (i) Given points $P(-3, 1)$ and $Q(3, 4)$, find the vector $\vec{PQ}$. (ii) Given point $P(-3, 1)$ and vector $\vec{PR} = \begin{pmatrix} 2 \\ -6 \end{pmatrix}$, find the coordinates of point $R$. (iii) Find the equation of the line $PQ$. The image only has the first two questions.

1. Problem Description

The problem consists of three parts.

(i) Given points $P(-3, 1)$ and $Q(3, 4)$, find the vector $\vec{PQ}$.

(ii) Given point $P(-3, 1)$ and vector $\vec{PR} = \begin{pmatrix} 2 \\ -6 \end{pmatrix}$, find the coordinates of point $R$.

(iii) Find the equation of the line $PQ$. The image only has the first two questions.

2. Solution Steps

(i) To find the vector $\vec{PQ}$, 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) To find the coordinates of point $R$, given point $P(-3, 1)$ and $\vec{PR} = \begin{pmatrix} 2 \\ -6 \end{pmatrix}$, 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:

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

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

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

3. Final Answer

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

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