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A-Level Pure 4 · Edexcel IAL WMA14 · Topic pack

Topic pack · Vectors แพ็กฝึกเฉพาะหัวข้อ · เวกเตอร์สามมิติ

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที 3-D vectors & lines, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

The vector $\mathbf{a}=2\mathbf{i}-\mathbf{j}+2\mathbf{k}$. (a) Find $|\mathbf{a}|$. (b) Write down the unit vector in the direction of $\mathbf{a}$.

Mark schemeshow ▾
(a) M1 $|\mathbf{a}|=\sqrt{2^2+(-1)^2+2^2}=\sqrt{9}$  A1 $=3$.
(b) B1 $\hat{\mathbf{a}}=\tfrac13\left(2\mathbf{i}-\mathbf{j}+2\mathbf{k}\right)$.
Q2. 4 marks

$\mathbf{a}=3\mathbf{i}+2\mathbf{j}-\mathbf{k}$ and $\mathbf{b}=2\mathbf{i}-\mathbf{j}+4\mathbf{k}$. (a) Find $\mathbf{a}\cdot\mathbf{b}$. (b) State what your answer tells you about $\mathbf{a}$ and $\mathbf{b}$. (c) Find $|\mathbf{b}|$.

Mark schemeshow ▾
(a) M1 $\mathbf{a}\cdot\mathbf{b}=6-2-4$  A1 $=0$.
(b) B1 $\mathbf{a}$ and $\mathbf{b}$ are perpendicular.
(c) B1 $|\mathbf{b}|=\sqrt{4+1+16}=\sqrt{21}$.
Q3. 4 marks

Find the angle between the vectors $\mathbf{a}=\mathbf{i}+2\mathbf{j}+2\mathbf{k}$ and $\mathbf{b}=4\mathbf{i}+3\mathbf{k}$, giving your answer in degrees to 1 decimal place.

Mark schemeshow ▾
M1 $\mathbf{a}\cdot\mathbf{b}=4+0+6=10$  B1 $|\mathbf{a}|=3$, $|\mathbf{b}|=5$  M1 $\cos\theta=\dfrac{10}{3\times 5}=\dfrac{2}{3}$  A1 $\theta=48.2^\circ$.
Q4. 5 marks

$\mathbf{a}=2\mathbf{i}+p\,\mathbf{j}-3\mathbf{k}$ and $\mathbf{b}=p\,\mathbf{i}+4\mathbf{j}+2\mathbf{k}$, where $p$ is a constant. Given that $\mathbf{a}$ and $\mathbf{b}$ are perpendicular, (a) find the value of $p$; (b) hence find $|\mathbf{a}|$.

Mark schemeshow ▾
(a) M1 $\mathbf{a}\cdot\mathbf{b}=2p+4p-6$  A1 $=6p-6$  M1 set $\mathbf{a}\cdot\mathbf{b}=0$  A1 $p=1$.
(b) B1 $|\mathbf{a}|=\sqrt{4+1+9}=\sqrt{14}$.
Q5. 5 marks

The points $A(1,\,2,\,-1)$ and $B(3,\,-2,\,5)$ are given. (a) Find a vector equation of the line through $A$ and $B$. (b) Show that the point $C(4,\,-4,\,8)$ lies on this line.

Mark schemeshow ▾
(a) M1 $\overrightarrow{AB}=\begin{pmatrix}2\\-4\\6\end{pmatrix}$  A1 simplify direction to $\begin{pmatrix}1\\-2\\3\end{pmatrix}$  A1 $\mathbf{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+t\begin{pmatrix}1\\-2\\3\end{pmatrix}$ (any correct form).
(b) M1 $x$: $1+t=4\Rightarrow t=3$; check the other two components with $t=3$  A1 $y$: $2-6=-4$ ✓ and $z$: $-1+9=8$ ✓ — all three components agree, so $C$ lies on the line.
Q6. 6 marks

The lines $l_1:\ \mathbf{r}=\begin{pmatrix}0\\1\\1\end{pmatrix}+s\begin{pmatrix}1\\1\\2\end{pmatrix}$ and $l_2:\ \mathbf{r}=\begin{pmatrix}3\\5\\4\end{pmatrix}+t\begin{pmatrix}1\\2\\-1\end{pmatrix}$. Show that $l_1$ and $l_2$ intersect, and find the coordinates of the point of intersection.

Mark schemeshow ▾
M1 equate components: $s=3+t$, $1+s=5+2t$, $1+2s=4-t$  M1 solve the first two: $1+(3+t)=5+2t$  A1 $t=-1$, $s=2$  M1 check the third component: $1+2(2)=5$ and $4-(-1)=5$  A1 consistent in all three components, so the lines intersect  A1 point $(2,\,3,\,5)$.
Q7. 6 marks

The lines $l_1:\ \mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+s\begin{pmatrix}2\\1\\3\end{pmatrix}$ and $l_2:\ \mathbf{r}=\begin{pmatrix}0\\1\\5\end{pmatrix}+t\begin{pmatrix}1\\-1\\2\end{pmatrix}$. Show that $l_1$ and $l_2$ are skew.

Mark schemeshow ▾
B1 $\begin{pmatrix}2\\1\\3\end{pmatrix}$ is not a scalar multiple of $\begin{pmatrix}1\\-1\\2\end{pmatrix}$, so the lines are not parallel  M1 equate $x$ and $y$: $1+2s=t$ and $s=1-t$  M1 solve: $1+2(1-t)=t\Rightarrow 3=3t$  A1 $t=1$, $s=0$  M1 check $z$: $2+3(0)=2$ but $5+2(1)=7$  A1 $2\ne 7$, so no intersection; not parallel and not intersecting → skew ∎.
Q8. 7 marks

The lines $l_1:\ \mathbf{r}=\begin{pmatrix}1\\3\\-2\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\2\end{pmatrix}$ and $l_2:\ \mathbf{r}=\begin{pmatrix}2\\1\\4\end{pmatrix}+\mu\begin{pmatrix}1\\2\\2\end{pmatrix}$. (a) Show that the point $A(7,\,0,\,4)$ lies on $l_1$. (b) Find the acute angle between $l_1$ and $l_2$, giving your answer in degrees to 1 decimal place.

Mark schemeshow ▾
(a) M1 $x$: $1+2\lambda=7\Rightarrow\lambda=3$; check $y$ and $z$ with $\lambda=3$  A1 $y$: $3-3=0$ ✓, $z$: $-2+6=4$ ✓ — all three components agree ∎.
(b) M1 $\begin{pmatrix}2\\-1\\2\end{pmatrix}\cdot\begin{pmatrix}1\\2\\2\end{pmatrix}=2-2+4=4$  B1 both direction vectors have magnitude $3$  M1 $\cos\theta=\dfrac{4}{3\times 3}$  A1 $\cos\theta=\dfrac{4}{9}$  A1 $\theta=63.6^\circ$.
Q9. 7 marks

The line $l$ has equation $\mathbf{r}=\begin{pmatrix}1\\0\\3\end{pmatrix}+t\begin{pmatrix}2\\-1\\2\end{pmatrix}$ and $P$ is the point $(5,\,1,\,4)$. (a) Find the coordinates of the point $F$ on $l$ that is closest to $P$. (b) Hence find the shortest distance from $P$ to $l$.

Mark schemeshow ▾
(a) M1 general point $Q(1+2t,\,-t,\,3+2t)$ on $l$  M1 $\overrightarrow{PQ}=\begin{pmatrix}2t-4\\-t-1\\2t-1\end{pmatrix}$  M1 perpendicularity: $\overrightarrow{PQ}\cdot\begin{pmatrix}2\\-1\\2\end{pmatrix}=0$, i.e. $(4t-8)+(t+1)+(4t-2)=9t-9=0$  A1 $t=1$  A1 $F(3,\,-1,\,5)$.
(b) M1 $\overrightarrow{PF}=\begin{pmatrix}-2\\-2\\1\end{pmatrix}$, distance $=\sqrt{4+4+1}$  A1 $=3$.
Q10. 8 marks

The points $A(1,\,2,\,3)$, $B(3,\,3,\,5)$ and $C(3,\,0,\,2)$ are given. (a) Find a vector equation of the line through $A$ and $B$. (b) Show that angle $BAC$ is a right angle. (c) Find the exact area of triangle $ABC$.

Mark schemeshow ▾
(a) M1 $\overrightarrow{AB}=\begin{pmatrix}2\\1\\2\end{pmatrix}$  A1 $\mathbf{r}=\begin{pmatrix}1\\2\\3\end{pmatrix}+t\begin{pmatrix}2\\1\\2\end{pmatrix}$.
(b) M1 $\overrightarrow{AC}=\begin{pmatrix}2\\-2\\-1\end{pmatrix}$ and $\overrightarrow{AB}\cdot\overrightarrow{AC}=4-2-2$  A1 $=0$, so $\overrightarrow{AB}\perp\overrightarrow{AC}$ and angle $BAC=90^\circ$ ∎.
(c) B1 $|\overrightarrow{AB}|=\sqrt{9}=3$  B1 $|\overrightarrow{AC}|=\sqrt{9}=3$  M1 area $=\tfrac12\,|\overrightarrow{AB}|\,|\overrightarrow{AC}|$ (right angle at $A$)  A1 $=\tfrac12\cdot 3\cdot 3=\tfrac{9}{2}$.