S4 E Maths Textbook Solutions >> New Syllabus Mathematics (NSM) 4 textbook solutions >>
Ex 5B
Please email me if you spot any mistakes or have any questions.
Solutions
Click to display or to hide
(a)
\begin{align*} {2 \choose 4} + {7 \choose 5} & = {9 \choose 9} \end{align*}
(b)
\begin{align*} {3 \choose -5} + {-4 \choose 1} & = {-1 \choose -4} \end{align*}
(c)
\begin{align*} {-9 \choose -2} + {0 \choose -8} + {-3 \choose 7} & = {-9 \choose -10} + {-3 \choose 7} \\ & = {-12 \choose -3} \end{align*}
(a)
\begin{align*} \textbf{a} + \textbf{b} & = {4 \choose 3} + {-7 \choose 2} \\ & = {-3 \choose 5} \\ \\ \textbf{b} + \textbf{a} & = {-7 \choose 2} + {4 \choose 3} \\ & = {-3 \choose 5} \\ \\ \text{Yes, } & \textbf{a} + \textbf{b} = \textbf{b} + \textbf{a} \end{align*}
(b)
\begin{align*} (\textbf{a} + \textbf{b}) + \textbf{c} & = {-3 \choose 5} + {1 \choose -5} \\ & = {-2 \choose 0} \\ \\ \textbf{a} + (\textbf{b} + \textbf{c}) & = {4 \choose 3} + \left( {-7 \choose 2} + {1 \choose -5} \right) \\ & = {4 \choose 3} + {-6 \choose -3} \\ & = {-2 \choose 0} \\ \\ \text{Yes, } & (\textbf{a} + \textbf{b}) + \textbf{c} = \textbf{a} + (\textbf{b} + \textbf{c}) \end{align*}
(a)
\begin{align*} \overrightarrow{LM} + \overrightarrow{MN} & = \overrightarrow{LN} \end{align*}
(b)
\begin{align*} \overrightarrow{PN} + \overrightarrow{LP} & = \overrightarrow{LP} + \overrightarrow{PN} \\ & = \overrightarrow{LN} \end{align*}
(c)
\begin{align*} \overrightarrow{LN} + \overrightarrow{NM} + \overrightarrow{MP} & = \overrightarrow{LM} + \overrightarrow{MP} \\ & = \overrightarrow{LP} \end{align*}
(a)
\begin{align*} {12 \choose -6} + {-12 \choose 6} & = {0 \choose 0} \end{align*}
(b)
\begin{align*} {5 \choose 7} + {-5 \choose -7} & = {0 \choose 0} \end{align*}
(c)
\begin{align*} {x \choose y} + {-x \choose -y} & = {0 \choose 0} \end{align*}
(a)
\begin{align*} {9 \choose 1} + {-9 \choose -1} & = {0 \choose 0} \end{align*}
(b)
\begin{align*} {-3 \choose -7} + {3 \choose 7} & = \textbf{0} \end{align*}
(c)
\begin{align*} {q \choose p} + {-q \choose -p} & = {0 \choose 0} \end{align*}
(a)
\begin{align*} \overrightarrow{AB} + \overrightarrow{BA} & = \textbf{0} \end{align*}
(b)
\begin{align*} \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RP} & = \overrightarrow{PR} + \overrightarrow{RP} \\ & = \textbf{0} \end{align*}
(c)
\begin{align*} \overrightarrow{MN} + \overrightarrow{LM} + \overrightarrow{NL} & = \overrightarrow{MN} + \overrightarrow{NL} + \overrightarrow{LM} \\ & = \overrightarrow{ML} + \overrightarrow{LM} \\ & = \textbf{0} \end{align*}
(a)
$$ -\textbf{q} + \textbf{p} $$
(b)
$$ -\textbf{p} + \textbf{q} $$
(c)
$$ -\textbf{a} + \textbf{b} $$
(d)
$$ \textbf{b} + \textbf{a} $$
(e)
$$ -\textbf{r} + \textbf{s} $$
(f)
$$ \textbf{r} + \textbf{s} $$
(g)
$$ -\textbf{n} - \textbf{m} $$
(h)
$$ -\textbf{m} + \textbf{n} $$
(a)
\begin{align*} {5 \choose 4} - {3 \choose 2} & = {2 \choose 2} \end{align*}
(b)
\begin{align*} {-1 \choose 3} - {-3 \choose -4} & = {-1 + 3 \choose 3 + 4} \\ & = {2 \choose 7} \end{align*}
(c)
\begin{align*} {2 \choose 3} + {5 \choose -2} - {7 \choose -3} & = {7 \choose 1} - {7 \choose -3} \\ & = {7 - 7 \choose 1 + 3} \\ & = {0 \choose 4} \end{align*}
(d)
\begin{align*} {4 \choose 7} - {-2 \choose 5} - {3 \choose -6} & = {4 + 2 \choose 7 - 5} - {3 \choose -6} \\ & = {6 \choose 2} - {3 \choose -6} \\ & = {6 - 3 \choose 2+ 6} \\ & = {3 \choose 8} \end{align*}
(a)
\begin{align*} \overrightarrow{PT} + \overrightarrow{TR} & = \overrightarrow{PR} \end{align*}
(b)
\begin{align*} \overrightarrow{SQ} + \overrightarrow{QR} & = \overrightarrow{SR} \end{align*}
(c)
\begin{align*} \overrightarrow{TR} + \overrightarrow{ST} & = \overrightarrow{ST} + \overrightarrow{TR} \\ & = \overrightarrow{SR} \end{align*}
(d)
\begin{align*} \overrightarrow{SQ} + \overrightarrow{QT} & = \overrightarrow{ST} \end{align*}
(e)
\begin{align*} \overrightarrow{SQ} + \overrightarrow{QR} + \overrightarrow{PS} & = \overrightarrow{SR} + \overrightarrow{PS} \\ & = \overrightarrow{PS} + \overrightarrow{SR} \\ & = \overrightarrow{PR} \end{align*}
(f)
\begin{align*} \overrightarrow{RQ} + \overrightarrow{QT} + \overrightarrow{TP} + \overrightarrow{PS} & = \overrightarrow{RT} + \overrightarrow{TP} + \overrightarrow{PS} \\ & = \overrightarrow{RP} + \overrightarrow{PS} \\ & = \overrightarrow{RS} \end{align*}
(a)
\begin{align*} \overrightarrow{RT} & = \overrightarrow{OS} \phantom{000} [\text{Opposite sides are parallel and equal in length}] \\ & = \textbf{s} \end{align*}
(b)
\begin{align*} \overrightarrow{ST} & = \overrightarrow{OR} \\ & = \textbf{r} \\ \\ \overrightarrow{TS} & = -\overrightarrow{ST} \\ & = -\textbf{r} \end{align*}
(c)
\begin{align*} \overrightarrow{OT} & = \overrightarrow{OR} + \overrightarrow{RT} \\ & = \textbf{r} + \textbf{s} \end{align*}
(d)
\begin{align*} \overrightarrow{RS} & = \overrightarrow{RO} + \overrightarrow{OS} \\ & = -\textbf{r} + \textbf{s} \end{align*}
(e)
\begin{align*} \overrightarrow{SR} & = -\overrightarrow{RS} \\ & = -(-\textbf{r} + \textbf{s}) \\ & = \textbf{r} - \textbf{s} \end{align*}
(a)
\begin{align*} \overrightarrow{RS} + \overrightarrow{ST} & = \overrightarrow{RT} \end{align*}
(b)
\begin{align*} \overrightarrow{RS} - \overrightarrow{RT} & = \overrightarrow{RS} + \overrightarrow{TR} \\ & = \overrightarrow{TR} + \overrightarrow{RS} \\ & = \overrightarrow{TS} \end{align*}
(c)
\begin{align*} \overrightarrow{RT} - \overrightarrow{RS} & = \overrightarrow{RT} + \overrightarrow{SR} \\ & = \overrightarrow{SR} + \overrightarrow{RT} \\ & = \overrightarrow{ST} \end{align*}
(d)
$$ \text{Cannot simplify} $$
(e)
\begin{align*} \overrightarrow{RS} - \overrightarrow{TS} & = \overrightarrow{RS} + \overrightarrow{ST} \\ & = \overrightarrow{RT} \end{align*}
(f)
\begin{align*} \overrightarrow{RS} + \overrightarrow{TR} - \overrightarrow{TU} & = \overrightarrow{TR} + \overrightarrow{RS} - \overrightarrow{TU} \\ & = \overrightarrow{TS} - \overrightarrow{TU} \\ & = \overrightarrow{TS} + \overrightarrow{UT} \\ & = \overrightarrow{UT} + \overrightarrow{TS} \\ & = \overrightarrow{US} \end{align*}
(a)
\begin{align*} {x \choose y} + {-3 \choose 2} & = {7 \choose -5} \\ {x - 3 \choose y + 2} & = {7 \choose -5} \\ \\ x - 3 & = 7 \\ x & = 7 + 3 \\ x & = 10 \\ \\ y + 2 & = - 5 \\ y & = -7 \end{align*}
(b)
\begin{align*} {3 \choose y} - {x \choose -8} & = {-6 \choose 9} \\ {3 - x \choose y + 8} & = {-6 \choose 9} \\ \\ 3 - x & = -6 \\ -x & = -6 - 3 \\ -x & = -9 \\ x & = 9 \\ \\ y + 8 & = 9 \\ y & = 9 - 8 \\ y & = 1 \end{align*}
(c)
\begin{align*} {y \choose 3} + {-4 \choose 2x} & = {6 \choose x} \\ {y - 4 \choose 3 + 2x} & = {6 \choose x} \\ \\ y - 4 & = 6 \\ y & = 6 + 4 \\ y & = 10 \\ \\ 3 + 2x & = x \\ 2x - x & = -3 \\ x & = -3 \end{align*}
(d)
\begin{align*} {2x \choose 5} - {y - 3 \choose -10} & = {4 \choose 3y} \\ {2x - (y - 3) \choose 5 + 10} & = {4 \choose 3y} \\ {2x - y + 3 \choose 15} & = {4 \choose 3y} \\ \\ 15 & = 3y \\ {15 \over 3} & = y \\ 5 & = y \\ \\ 2x - y + 3 & = 4 \\ 2x & = 4 - 3 + y \\ 2x & = 1 + y \\ 2x & = 1 + (5) \\ 2x & = 6 \\ x & = {6 \over 2} \\ x & = 3 \end{align*}
(a)(i)
(a)(ii)
(a)(iii)
(a)(iv)
(b)
$$ \text{Yes. The resulting vectors have the same magnitude and the same direction.} $$
(c)
$$ \text{Yes. The resulting vectors have the same magnitude and the same direction.} $$
(a)(i)
\begin{align*} \overrightarrow{QR} & = \overrightarrow{PS} \phantom{000} [\text{Opposite sides of parallelogram}] \\ \\ \overrightarrow{PQ} + \overrightarrow{PS} & = \overrightarrow{PQ} + \overrightarrow{QR} \\ & = \overrightarrow{PR} \end{align*}
(a)(ii)
\begin{align*} \overrightarrow{RO} - \overrightarrow{QO} & = \overrightarrow{RO} + \overrightarrow{OQ} \\ & = \overrightarrow{RQ} \end{align*}
(a)(iii)
\begin{align*} \overrightarrow{PR} - \overrightarrow{SR} + \overrightarrow{SQ} & = \overrightarrow{PR} + \overrightarrow{RS} + \overrightarrow{SQ} \\ & = \overrightarrow{PS} + \overrightarrow{SQ} \\ & = \overrightarrow{PQ} \end{align*}
(b)(i)
\begin{align*} \overrightarrow{SR} & = \overrightarrow{PQ} \\ & = \textbf{a} \end{align*}
(b)(ii)
\begin{align*} \overrightarrow{PR} & = \overrightarrow{PQ} + \overrightarrow{QR} \\ & = \textbf{a} + \overrightarrow{PS} \\ & = \textbf{a} + \textbf{b} \end{align*}
(b)(iii)
\begin{align*} \overrightarrow{SQ} & = \overrightarrow{SP} + \overrightarrow{PQ} \\ & = -\overrightarrow{PS} + \overrightarrow{PQ} \\ & = -\textbf{b} + \textbf{a} \end{align*}
(a)
\begin{align*} \overrightarrow{SK} + \overrightarrow{KS} & = \textbf{0} \\ \\ \textbf{u} = & \phantom{.} \overrightarrow{KS} \end{align*}
(b)
\begin{align*} \overrightarrow{SP} + \overrightarrow{PQ} + \textbf{u} & = \textbf{0} \\ \overrightarrow{SQ} + \textbf{u} & = \textbf{0} \\ \\ \textbf{u} = & \phantom{.} \overrightarrow{QS} \end{align*}
(c)
\begin{align*} \overrightarrow{PS} + \overrightarrow{SK} + \overrightarrow{KR} & = \textbf{u} \\ \overrightarrow{PK} + \overrightarrow{KR} & = \textbf{u} \\ \overrightarrow{PR} & = \textbf{u} \end{align*}
(d)
\begin{align*} \overrightarrow{PK} + (-\overrightarrow{SK}) & = \textbf{u} \\ \overrightarrow{PK} + \overrightarrow{KS} & = \textbf{u} \\ \overrightarrow{PS} & = \textbf{u} \end{align*}
(e)
\begin{align*} \overrightarrow{PS} + (-\overrightarrow{RS}) & = \textbf{u} \\ \overrightarrow{PS} + \overrightarrow{SR} & = \textbf{u} \\ \overrightarrow{PR} & = \textbf{u} \end{align*}
(f)
\begin{align*} \overrightarrow{PQ} + \overrightarrow{QR} + (-\overrightarrow{PR}) & = \textbf{u} \\ \overrightarrow{PR} + (-\overrightarrow{PR}) & = \textbf{u} \\ \overrightarrow{PR} + \overrightarrow{RP} & = \textbf{u} \\ \textbf{0} & = \textbf{u} \end{align*}