点 A 通过向量 $\begin{pmatrix}-5\\4\end{pmatrix}$ 平移到点 B。 点 B 的位置向量为 $\begin{pmatrix}3\\8\end{pmatrix}$。 求 $|\overrightarrow{OA}|$。 答案 ... kg P 是直线 AB 上的一个点，其坐标为 (k, 10)。
求 k 的值。 答案 k = ... 用 p 和 r 表示以下各项。请将你的答案写成最简形式。 $\overrightarrow{OQ}$ $\overrightarrow{PQ}$ $\overrightarrow{OB}$ 解释为什么点 O、B 和 Q 不共线。

1. Find the position vector of A from the translation to B.

A is translated to B by $\begin{pmatrix}-5\\4\end{pmatrix}$, so
$\overrightarrow{OB}=\overrightarrow{OA}+\begin{pmatrix}-5\\4\end{pmatrix}.$
Given $\overrightarrow{OB}=\begin{pmatrix}3\\8\end{pmatrix}$,
$\overrightarrow{OA}=\begin{pmatrix}3\\8\end{pmatrix}-\begin{pmatrix}-5\\4\end{pmatrix}=\begin{pmatrix}8\\4\end{pmatrix}.$

2. Calculate the magnitude of $\overrightarrow{OA}$.

$|\overrightarrow{OA}|=\sqrt{8^2+4^2}=\sqrt{80}=4\sqrt{5}.$

1. Find the coordinates of A using the translation from A to B.

Point A is translated to point B by $\begin{pmatrix}-5\\4\end{pmatrix}$, so
$\overrightarrow{OB}=\overrightarrow{OA}+\begin{pmatrix}-5\\4\end{pmatrix}.$
Given $\overrightarrow{OB}=\begin{pmatrix}3\\8\end{pmatrix}$,
$\overrightarrow{OA}=\begin{pmatrix}3\\8\end{pmatrix}-\begin{pmatrix}-5\\4\end{pmatrix}=\begin{pmatrix}8\\4\end{pmatrix}.$
So $A=(8,4)$ and $B=(3,8)$.

2. Use collinearity on AB to find the x-coordinate when y = 10.

The vector from A to B is
$\overrightarrow{AB}=\begin{pmatrix}3-8\\8-4\end{pmatrix}=\begin{pmatrix}-5\\4\end{pmatrix}.$
Since $P$ lies on line $AB$, write
$P=A+t\overrightarrow{AB}=\begin{pmatrix}8\\4\end{pmatrix}+t\begin{pmatrix}-5\\4\end{pmatrix}.$
Using the y-coordinate of $P$,
$4+4t=10$
$t=\frac{6}{4}=\frac{3}{2}.$
Hence
$k=8-5\left(\frac{3}{2}\right)=8-\frac{15}{2}=\frac{1}{2}.$

1. Express $\overrightarrow{OQ}$ as the sum of two known vectors along the path $O \to R \to Q$.

$\overrightarrow{OQ}=\overrightarrow{OR}+\overrightarrow{RQ}$

$=4\mathbf{r}+(7\mathbf{p}+4\mathbf{r})$

$=7\mathbf{p}+8\mathbf{r}$

1. Find the vector from P to Q by subtracting the position vector of P from the position vector of Q.

From 6(b)(i),
$\overrightarrow{OQ}=7\mathbf{p}+8\mathbf{r}$ and $\overrightarrow{OP}=3\mathbf{p}$.

$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}$
$=(7\mathbf{p}+8\mathbf{r})-3\mathbf{p}$
$=4\mathbf{p}+8\mathbf{r}$

1. Find the vector from P to R.

$\overrightarrow{PR}=\overrightarrow{OR}-\overrightarrow{OP}$
$=4\mathbf{r}-3\mathbf{p}$

2. Use the ratio $PB:BR=3:2$ to express $\overrightarrow{OB}$.

Since $PB:BR=3:2$,
$\overrightarrow{PB}=\frac{3}{5}\overrightarrow{PR}$
$\overrightarrow{OB}=\overrightarrow{OP}+\overrightarrow{PB}$
$=3\mathbf{p}+\frac{3}{5}(4\mathbf{r}-3\mathbf{p})$
$=3\mathbf{p}+\frac{12}{5}\mathbf{r}-\frac{9}{5}\mathbf{p}$
$=\frac{6}{5}\mathbf{p}+\frac{12}{5}\mathbf{r}$

1. Compare the position vectors of B and Q from O.

From earlier parts,
$\overrightarrow{OB}=\frac{6}{5}\mathbf{p}+\frac{12}{5}\mathbf{r},\qquad \overrightarrow{OQ}=7\mathbf{p}+8\mathbf{r}.$
For $O$, $B$ and $Q$ to be collinear, $\overrightarrow{OB}$ must be a scalar multiple of $\overrightarrow{OQ}$.

But
$\frac{\frac{6}{5}}{7}\ne \frac{\frac{12}{5}}{8}.$
So $\overrightarrow{OB}$ is not parallel to $\overrightarrow{OQ}$.