$\overrightarrow{AB}=\begin{pmatrix}-3\\9\end{pmatrix},\ \overrightarrow{BC}=\begin{pmatrix}2\\5\end{pmatrix},\ \overrightarrow{CD}=\begin{pmatrix}d\\4\end{pmatrix}.$ 已知 $\overrightarrow{CD}$ 平行于 $\overrightarrow{AB}$，求 $d$ 的值。 答 求 $|\overrightarrow{AC}|$。 答 ... 单位 PQR 是一个三角形，M 是 PQ 的中点。已知 $\overrightarrow{PQ}=a$，$\overrightarrow{PR}=b$，$\overrightarrow{QT}=h\,\overrightarrow{QR}$ 且 $\overrightarrow{RS}=k\,\overrightarrow{RM}$，其中 k 和 h 是常数。 用 k、a 和 b 表示 $\overrightarrow{PS}$。 答 用 h、a 和 b 表示 $\overrightarrow{PT}$。 答 由此证明，若 P、S 和 T 共线，则 $hk+2=2(h+k)$。 答 已知 $h=\dfrac{1}{3}$ 且 $k=\dfrac{4}{5}$，求 PS : ST 的比值。 答 ... : ...

1. Set up the scalar multiple relationship for parallel vectors

Since $\overrightarrow{CD}$ is parallel to $\overrightarrow{AB}$,
$\begin{pmatrix}d\\4\end{pmatrix}=\lambda \begin{pmatrix}-3\\9\end{pmatrix}$
for some scalar $\lambda$.

2. Compare components and find $d$

Compare the second components:
$4=9\lambda$
$\lambda=\frac{4}{9}$
Compare the first components:
$d=-3\lambda=-3\left(\frac{4}{9}\right)=-\frac{4}{3}$

1. Find the vector $\overrightarrow{AC}$ by adding $\overrightarrow{AB}$ and $\overrightarrow{BC}$.

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$
$=\begin{pmatrix}-3\\9\end{pmatrix}+\begin{pmatrix}2\\5\end{pmatrix}=\begin{pmatrix}-1\\14\end{pmatrix}$

2. Find the magnitude of $\overrightarrow{AC}$.

$|\overrightarrow{AC}|=\sqrt{(-1)^2+14^2}$
$=\sqrt{197}$

1. Express $\overrightarrow{RM}$ in terms of $a$ and $b$.

Since $M$ is the midpoint of $PQ$,

$\overrightarrow{PM}=\dfrac{1}{2}\overrightarrow{PQ}=\dfrac{1}{2}a$.

Also, $\overrightarrow{RP}=-\overrightarrow{PR}=-b$.

Hence

$\overrightarrow{RM}=\overrightarrow{RP}+\overrightarrow{PM}=-b+\dfrac{1}{2}a.$

2. Use $\overrightarrow{PS}=\overrightarrow{PR}+\overrightarrow{RS}$ and substitute $\overrightarrow{RS}=k\overrightarrow{RM}$.

$\overrightarrow{PS}=\overrightarrow{PR}+\overrightarrow{RS}$

$=b+k\left(-b+\dfrac{1}{2}a\right)$

$=b-kb+\dfrac{k}{2}a$

$=\left(1-k\right)b+\dfrac{k}{2}a.$

1. Express $\overrightarrow{QR}$ in terms of $a$ and $b$.

Since $\overrightarrow{PQ}=a$ and $\overrightarrow{PR}=b$,
$\overrightarrow{QR}=\overrightarrow{QP}+\overrightarrow{PR}=-a+b=b-a.$

2. Use $\overrightarrow{PT}=\overrightarrow{PQ}+\overrightarrow{QT}$ and substitute $\overrightarrow{QT}=h\overrightarrow{QR}$.

$\overrightarrow{PT}=\overrightarrow{PQ}+\overrightarrow{QT}$
$=a+h(b-a)$
$=a+hb-ha$
$=(1-h)a+hb.$

1. Use collinearity to equate vector components and eliminate the scalar multiple.

From earlier parts,
$\overrightarrow{PS}=\frac{k}{2}a+(1-k)b$
and
$\overrightarrow{PT}=(1-h)a+hb.$
Since $P$, $S$ and $T$ are collinear, let
$\overrightarrow{PS}=\lambda \overrightarrow{PT}.$
Comparing coefficients of $a$ and $b$,
$\frac{k}{2}=\lambda(1-h) \quad \text{and} \quad 1-k=\lambda h.$
Hence
$\lambda=\frac{k}{2(1-h)}=\frac{1-k}{h}.$
So
$hk=2(1-h)(1-k).$
Expanding,
$hk=2-2h-2k+2hk.$
Therefore
$hk+2=2h+2k=2(h+k).$

1. Substitute $k=\dfrac45$ into the expression for $\overrightarrow{PS}$.

From 8(b)(i),
$\overrightarrow{PS}=(1-k)b+\frac{k}{2}a.$
When $k=\dfrac45$,
$\overrightarrow{PS}=\left(1-\frac45\right)b+\frac{1}{2}\left(\frac45\right)a$
$=\frac15 b+\frac25 a.$

2. Substitute $h=\dfrac13$ into the expression for $\overrightarrow{PT}$ and compare with $\overrightarrow{PS}$.

From 8(b)(ii),
$\overrightarrow{PT}=(1-h)a+hb.$
When $h=\dfrac13$,
$\overrightarrow{PT}=\left(1-\frac13\right)a+\frac13 b$
$=\frac23 a+\frac13 b.$
Now compare:
$\overrightarrow{PS}=\frac25a+\frac15b=\frac35\left(\frac23a+\frac13b\right).$
So
$\overrightarrow{PS}=\frac35\overrightarrow{PT}.$
Hence $PS:PT=3:5$, so
$ST=PT-PS=5-3=2\text{ parts}.$
Therefore
$PS:ST=3:2.$