Sec 4 E Maths: Vectors, practice questions & worked solutions

$$\begin{aligned}\sqrt{m^2+3.6^2}&=3.9\\m^2+3.6^2&=3.9^2\\m^2&=2.25\\m&=1.5\end{aligned}$$

Answer: $m=1.5$

(a)

$$\begin{aligned}\left|\vect{PR}\right|&=\sqrt{8^2+(-10)^2}\\[4pt]\left|\vect{PR}\right|&=2\sqrt{41}\text{ units}\end{aligned}$$

OR
$\left|\vect{PR}\right|=12.8$ units

(b)

Given $\vect{PR}=\begin{pmatrix}8\\-10\end{pmatrix}$,

$$\begin{aligned}\text{Gradient}&=\dfrac{-10}8\\[4pt]&=-\dfrac54\end{aligned}$$

Equation of $PR$, $y=mx+c$

$$y=-\dfrac54x+12$$

Answer: (a) $2\sqrt{41}$ units or $12.8$ units (b) $y=-\dfrac54x+12$

(a)

$$\begin{aligned}|\overrightarrow{QR}|&=\sqrt{8^2+6^2}\\&=\sqrt{100}\\&=10\text{ units}\end{aligned}$$

(b)

$$\begin{aligned}\overrightarrow{ST}&=\overrightarrow{SO}+\overrightarrow{OT}\\&=\begin{pmatrix}-1\\3\end{pmatrix}+\begin{pmatrix}5\\w\end{pmatrix}\\&=\begin{pmatrix}4\\3+w\end{pmatrix}\end{aligned}$$

(c)

$$\begin{aligned}\begin{pmatrix}4\\3+w\end{pmatrix}&=k\begin{pmatrix}1\\4\end{pmatrix}\\&=\begin{pmatrix}k\\4k\end{pmatrix}\\k&=4\\3+w&=4(4)\\w&=16-3\\&=13\end{aligned}$$

Answer: (a) 10 units  (b) $\begin{pmatrix}4\\3+w\end{pmatrix}$  (c) $w=13$

(a)

$$\begin{aligned}\vv{RS}&=-\mathbf r+\mathbf s\\[6pt]\vv{OJ}&=\vv{OR}+\vv{RJ}\\[6pt]&=\mathbf r+\frac35\,\vv{RS}\\[6pt]&=\mathbf r+\frac35(\mathbf s-\mathbf r)\\[6pt]&=\frac25\mathbf r+\frac35\mathbf s=\frac15(2\mathbf r+3\mathbf s)\text{ \textit{(shown)}}\end{aligned}$$

(b)

$$\begin{aligned}\vv{RX}&=\frac15(\mathbf r+9\mathbf s)\\[6pt]\vv{OX}-\vv{OR}&=\frac15\mathbf r+\frac95\mathbf s\\[6pt]\vv{OX}&=\frac15\mathbf r+\frac95\mathbf s+\mathbf r\\[6pt]&=\frac65\mathbf r+\frac95\mathbf s\\[6pt]\vv{OX}&=3\left[\frac25\mathbf r+\frac35\mathbf s\right]\\[6pt]\vv{OX}&=3\,\vv{OJ}\end{aligned}$$

Since the vectors $\vv{OX}$ and $\vv{OJ}$ differ by a constant of 3 and have a common point $O$, it means that the vectors are parallel.

This further implies that $OX$ and $OJ$ are parallel lines.

Thus, the points $O$, $J$ and $X$ lie on the same straight line.

Answer: (a) $\overrightarrow{OJ}=\mathbf r+\tfrac35(\mathbf s-\mathbf r)=\tfrac15(2\mathbf r+3\mathbf s)$   (b) $\overrightarrow{OX}=\tfrac65\mathbf r+\tfrac95\mathbf s=3\overrightarrow{OJ}$, so $OX$ is parallel to $OJ$ with the common point $O$: $O$, $J$ and $X$ lie on a straight line

(a)(i)

$$\begin{aligned}\overrightarrow{OF}&=\overrightarrow{OC}+\overrightarrow{CF}\\&=8\mathbf q-2\mathbf q+2\mathbf p\\&=6\mathbf q+2\mathbf p\end{aligned}$$

(a)(ii)

$$\begin{aligned}\overrightarrow{OD}&=8\mathbf q-2\mathbf p-6\mathbf q\\&=2\mathbf q-2\mathbf p\\\overrightarrow{AD}&=-3\mathbf p-2\mathbf p+2\mathbf q\\&=2\mathbf q-5\mathbf p\end{aligned}$$

Alternative method:

$$\begin{aligned}\overrightarrow{AD}&=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}\\&=5\mathbf q+3\mathbf q-3\mathbf p-2\mathbf p-6\mathbf q\\&=2\mathbf q-5\mathbf p\end{aligned}$$

(b)

Since $\dfrac32\overrightarrow{OD}=\overrightarrow{BC}$, $\overrightarrow{OD}$ is parallel to $\overrightarrow{BC}$, and magnitude of $\overrightarrow{BC}$ is $\dfrac32$ times that of $\overrightarrow{OD}$.

(c)

$\dfrac8{13}$

Answer: (a)(i) $6\mathbf q+2\mathbf p$ (ii) $2\mathbf q-5\mathbf p$ (b) Parallel, $|BC|=\dfrac32|OD|$ (c) $\dfrac8{13}$

(a)(i)

$$\begin{aligned}\vect{PL}&=\frac13\vect{PT}\\\vect{PL}&=\frac13(m-p)\end{aligned}$$

(a)(ii)

$$\begin{aligned}\vect{KL}&=\vect{KO}+\vect{OP}+\vect{PL}\\\vect{KL}&=-\frac23m+p+\frac13m-\frac13p\\\vect{KL}&=\frac23p-\frac13m\end{aligned}$$

(b)

$$\begin{aligned}\vect{OM}&=\vect{OK}+\vect{KM}\\\vect{OM}&=\frac23m-\frac23m+\frac43p\\\vect{OM}&=\frac43p\\\vect{OM}&=\frac43\vect{OP}\end{aligned}$$

Since $\vect{OM}=\dfrac43\vect{OP}$ and O is a common point, therefore M lies on OP extended.

(c)

$$\begin{aligned}\frac{\textit{Area KTL}}{\textit{Area OTP}}&=\frac{\textit{Area KTL}}{\textit{Area KPT}}\times\frac{\textit{Area KPT}}{\textit{Area OTP}}\\[4pt]\frac{\textit{Area KTL}}{\textit{Area OTP}}&=\frac23\times\frac13=\frac29\end{aligned}$$

Answer: (a)(i) $\frac13(\mathbf m-\mathbf p)$ (ii) $\frac23\mathbf p-\frac13\mathbf m$ (c) $2:9$

(a)(i)

$\overrightarrow{BF}=-\mathbf a+\mathbf c$

(a)(ii)

$$\begin{aligned}\overrightarrow{AE}&=\mathbf c+\frac13(\mathbf c-\mathbf a)\\&=\frac13(4\mathbf c-\mathbf a)\end{aligned}$$

(a)(iii)

$\overrightarrow{OD}=\dfrac12\mathbf a+\mathbf c$

(b)

$$\begin{aligned}\overrightarrow{OE}&=\overrightarrow{OA}+\overrightarrow{AE}\\&=\mathbf a+\frac13(4\mathbf c-\mathbf a)\\&=\frac23\mathbf a+\frac43\mathbf c\\&=\frac43\left(\frac12\mathbf a+\mathbf c\right)\end{aligned}$$

$\overrightarrow{OE}=\dfrac43\overrightarrow{OD}\Rightarrow\overrightarrow{OE}\parallel\overrightarrow{OD}$

$\overrightarrow{OE}$ and $\overrightarrow{OD}$ share a common point $O$ and therefore points $O$, $D$ and $E$ lie on the same straight line.

OR

$$\begin{aligned}\overrightarrow{DE}&=\overrightarrow{DB}+\overrightarrow{BE}\\&=\frac12\mathbf a+\frac13\overrightarrow{BF}\\&=\frac12\mathbf a+\frac13(-\mathbf a+\mathbf c)\\&=\frac16\mathbf a+\frac13\mathbf c\\&=\frac13\left(\frac12\mathbf a+\mathbf c\right)\end{aligned}$$

$\overrightarrow{DE}=\dfrac13\overrightarrow{OD}\Rightarrow\overrightarrow{DE}\parallel\overrightarrow{OD}$

$\overrightarrow{DE}$ and $\overrightarrow{OD}$ share a common point $D$ and therefore points $O$, $D$ and $E$ lie on the same straight line.

(c)

$$\begin{aligned}&\frac{\text{area of triangle }CDE}{\text{area of parallelogram }OABC}\\&=\frac{\text{Area of }\triangle CDE}{\text{Area of }\triangle CDO}\times\frac{\text{Area of }\triangle CDO}{\text{Area of parallelogram }OABC}\\&=\frac13\times\frac14\\&=\frac1{12}\end{aligned}$$

Answer: (a)(i) $\mathbf c-\mathbf a$   (ii) $\tfrac13(4\mathbf c-\mathbf a)$   (iii) $\tfrac12\mathbf a+\mathbf c$   (b) yes: $\overrightarrow{OE}=\tfrac23\mathbf a+\tfrac43\mathbf c=\tfrac43\overrightarrow{OD}$ with the common point $O$   (c) $\dfrac1{12}$

(a)(i)

$$\begin{aligned}\vect{BC}&=\vect{OC}-\vect{OB}\\&=\underline q+\dfrac16(2\underline p)-2\underline p\\&=\underline q-\dfrac53\underline p\end{aligned}$$

(a)(ii)

$$\begin{aligned}\vect{XC}&=\vect{XB}+\vect{BC}\\&=\vect{DC}+\vect{BC}\\&=\dfrac16(2\underline p)+\underline q-\dfrac53\underline p\\&=\dfrac13\underline p+\underline q-\dfrac53\underline p\\&=\underline q-\dfrac43\underline p\end{aligned}$$

(a)(iii)

area $XBCD:$ area $OBCD$

$$\begin{aligned}&=\dfrac13:\dfrac12\left(2+\dfrac13\right)\\&=2:3\left(\dfrac73\right)\\&=2:7\end{aligned}$$

or

$$\begin{aligned}&=XB\times h:\dfrac12(DC+OB)h\\&=XB:\dfrac12(DC+OB)\\&=1:\dfrac12(1+6)\\&=1:\dfrac72\\&=2:7\end{aligned}$$

(a)(iv)

$$\begin{aligned}\vect{OY}&=\vect{OB}+\vect{BY}\\h\,\underline q&=2\underline p+k\underline q-\dfrac{5k}3\underline p\\h\,\underline q&=\left(2-\dfrac{5k}3\right)\underline p+k\underline q\\h=k,\quad2-\dfrac{5k}3&=0\\h=\dfrac65=1\dfrac15,\quad k&=\dfrac65=1\dfrac15\end{aligned}$$

(b)(i)

$$\begin{aligned}\vect{OT}&=\begin{pmatrix}-2\\6\end{pmatrix}\\\vect{OS}&=\vect{OT}+\vect{TS}\\&=\begin{pmatrix}-2\\6\end{pmatrix}-\begin{pmatrix}-7\\9\end{pmatrix}\\&=\begin{pmatrix}5\\-3\end{pmatrix}\end{aligned}$$

(b)(ii)

$$\begin{aligned}\text{Magnitude}&=\sqrt{(-7)^2+9^2}\\&=\sqrt{130}\approx11.401=11.4\text{ units (3 s.f.)}\end{aligned}$$

Answer: (a)(ii) $\mathbf q-\dfrac43\mathbf p$ (a)(iii) $2:7$ (a)(iv) $h=k=1.2$ (b)(i) $\begin{pmatrix}5\\-3\end{pmatrix}$ (b)(ii) 11.4

(a)(i)

$$\begin{aligned}\overrightarrow{AC}&=4\overrightarrow{AD}\\&=4\mathbf a\\\overrightarrow{BC}&=\overrightarrow{AC}+\overrightarrow{BA}\\&=4\mathbf a-3\mathbf b\end{aligned}$$

(a)(ii)

$$\begin{aligned}\overrightarrow{AE}&=\overrightarrow{AB}+\overrightarrow{BE}\\&=3\mathbf b+\frac12(4\mathbf a-3\mathbf b)\\&=2\mathbf a+\frac32\mathbf b\\\overrightarrow{AF}&=\frac25\overrightarrow{AE}\\&=\frac45\mathbf a+\frac35\mathbf b\end{aligned}$$

(a)(iii)

$$\begin{aligned}\overrightarrow{BF}&=\overrightarrow{BA}+\overrightarrow{AF}\\&=-3\mathbf b+\left(\frac45\mathbf a+\frac35\mathbf b\right)\\&=\frac45\mathbf a-\frac{12}5\mathbf b\end{aligned}$$

(b)

$$\begin{aligned}\overrightarrow{BF}&=\frac45\mathbf a-\frac{12}5\mathbf b=\frac45(\mathbf a-3\mathbf b)\\\overrightarrow{BD}&=\overrightarrow{BA}+\overrightarrow{AD}\\&=-3\mathbf b+\mathbf a\\\overrightarrow{BF}&=\frac45\overrightarrow{BD}\end{aligned}$$

The points $B$, $F$ and $D$ are collinear with common point $B$.

Hence, $BF$ produced will meet $AC$ at $D$.

(c)(i)

$$\begin{aligned}\frac{\text{area of }\triangle ABF}{\text{area of }\triangle ABE}&=\frac{AF}{AE}\\&=\frac25\end{aligned}$$

(c)(ii)

$$\begin{aligned}\frac{\text{area of }\triangle BEF}{\text{area of }\triangle ABC}&=\frac{\text{area of }\triangle BEF}{\text{area of }\triangle ABE}\times\frac{\text{area of }\triangle ABE}{\text{area of }\triangle ABC}\\&=\frac{FE}{AE}\times\frac{BE}{BC}\\&=\frac35\times\frac12\\&=\frac3{10}\end{aligned}$$

Answer: (a)(i) $4\mathbf a-3\mathbf b$ (ii) $\dfrac45\mathbf a+\dfrac35\mathbf b$ (iii) $\dfrac45\mathbf a-\dfrac{12}5\mathbf b$ (b) Yes (c)(i) $\dfrac25$ (ii) $\dfrac3{10}$

(a)(i)

$$\begin{aligned}\sqrt{(2p)^2+(5p)^2}&=\dfrac{5\sqrt{29}}2\\\sqrt{29p^2}&=\dfrac{5\sqrt{29}}2\\29p^2&=\dfrac{25\times29}4\\p&=\pm\sqrt{\dfrac{25\times29}4\times\dfrac1{29}}\\&=\pm\dfrac52\end{aligned}$$

(a)(ii)

$$\begin{aligned}\vect{XZ}&=\begin{pmatrix}6\\-2\end{pmatrix}-\begin{pmatrix}-6\\2\end{pmatrix}\\&=\begin{pmatrix}12\\-4\end{pmatrix}\end{aligned}$$

$$\begin{aligned}\vect{XY}&=\begin{pmatrix}3\\1\end{pmatrix}-\begin{pmatrix}-6\\2\end{pmatrix}&\text{or}&&\vect{YZ}&=\begin{pmatrix}6\\-2\end{pmatrix}-\begin{pmatrix}3\\1\end{pmatrix}\\&=\begin{pmatrix}9\\-1\end{pmatrix}&&&&=\begin{pmatrix}3\\-3\end{pmatrix}\end{aligned}$$

Since $\vect{XZ}\ne k\vect{XY}$ or $\vect{XZ}\ne k\vect{YZ}$,
or Since $\begin{pmatrix}12\\-4\end{pmatrix}\ne k\begin{pmatrix}9\\-1\end{pmatrix}$,
or since $\begin{pmatrix}12\\-4\end{pmatrix}\ne k\begin{pmatrix}3\\-3\end{pmatrix}$,
$Y(3,1)$ does not lie on this line.

(b)(i)(i)

$$\begin{aligned}\vect{CD}&=\vect{OD}-\vect{OC}\\&=2\mathbf b-\dfrac32\mathbf a\end{aligned}$$

(b)(i)(ii)

$$\begin{aligned}\vect{ON}&=\vect{OC}+\vect{CN}\\&=\dfrac32\mathbf a+\dfrac37\left(2\mathbf b-\dfrac32\mathbf a\right)\\&=\dfrac67(\mathbf a+\mathbf b)\end{aligned}$$

OR

$$\begin{aligned}\vect{ON}&=\vect{OD}+\vect{DN}\\&=2\mathbf b+\dfrac47\left(-2\mathbf b+\dfrac32\mathbf a\right)\\&=\dfrac67(\mathbf a+\mathbf b)\end{aligned}$$

(b)(ii)(i)

$2:3$

(b)(ii)(ii)

$$\begin{aligned}\dfrac{\text{area of }\triangle OAB}{\text{area of }\triangle OCD}&=\dfrac{\text{area of }\triangle OAB}{\text{area of }\triangle OCB}\times\dfrac{\text{area of }\triangle OCB}{\text{area of }\triangle OCD}\\&=\dfrac23\times\dfrac12\\&=\dfrac13\end{aligned}$$

Ratio $=1:3$
OR

$$\begin{aligned}\dfrac{\text{area of }\triangle OAB}{\text{area of }\triangle OCD}&=\dfrac{\frac12\times OA\times OB\times\sin\angle COD}{\frac12\times OC\times OD\times\sin\angle COD}\\&=\dfrac{\frac12\times\frac23OC\times\frac12OD\times\sin\angle COD}{\frac12\times OC\times OD\times\sin\angle COD}\\&=\dfrac13\end{aligned}$$

Ratio $=1:3$

Answer: (a)(i) $p=\pm\dfrac52$ (a)(ii) $Y$ does not lie on the line (b)(i)(a) $2\mathbf b-\dfrac32\mathbf a$ (b)(i)(b) $\dfrac67(\mathbf a+\mathbf b)$ (b)(ii)(a) $2:3$ (b)(ii)(b) $1:3$