Question 5:
Solutions by scale drawing are not accepted.
Diagram 5 shows two points on a Cartesian plane.
(a) $$
\text { Determine } \overrightarrow{A B} \text {. }
$$
[1 mark]
(b) $$ \text { Point } C \text { lies on the same Cartesian plane. } $$
$$ \text { It is given that } \overrightarrow{O C}=(m+2) \underline{i}+12 \underline{j} \text {, such that } m \text { is a constant. } $$
(i) $$ \text { If } \overleftrightarrow{O C} \text { is parallel to } \overrightarrow{A B} \text {, } $$
$$ \text { find the value of } m \text { by using vector’s arithmetic operations. } $$
(ii) $$ \text { Hence, determine the unit vector in the direction of } \overrightarrow{O C} \text {. } $$
$$ \text { State your answer in the form of column vector. } $$
[5 marks]
Answer:
(a)(i) $$ \begin{aligned} & \overrightarrow{A B}=\overrightarrow{A O}+\overrightarrow{O B} \\ & \overrightarrow{A B}=-\overrightarrow{O A}+\overrightarrow{O B} \\ & \overrightarrow{A B}=-(6 \underline{i}-11 \underline{j})+(-4 \underline{i}+13 \underline{j}) \\ & \overrightarrow{A B}=-6 \underline{i}+11 \underline{j}-4 \underline{i}+13 \underline{j} \\ & \overrightarrow{A B}=-10 \underline{i}+24 \underline{j} \end{aligned} $$
(b)(i) $$ \overrightarrow{O C} \| \overrightarrow{A B} $$
$$ \begin{aligned} \overrightarrow{O C} & =k \overrightarrow{A B} \\ (m+2) \underline{i}+12 \underline{j} & =k(-10 \underline{i}+24 \underline{j}) \\ (m+2) \underline{i}+12 \underline{j} & =-10 k \underline{i}+24 k \underline{j} \end{aligned} $$
$$ \begin{aligned} j: 24 k & =12 \\ k & =\frac{12}{24} \\ k & =\frac{1}{2} \end{aligned} $$
$$ \begin{aligned} \underline{i}: m+2 & =-10 k \\ m+2 & =-10\left(\frac{1}{2}\right) \\ m & =-5-2 \\ m & =-7 \end{aligned} $$
$$ \therefore m=-7 $$
(b)(ii) $$ \begin{aligned} & \overrightarrow{O C}=(-7+2) \underline{i}+12 \underline{j} \\ & \overrightarrow{O C}=-5 \underline{i}+12 \underline{j} \end{aligned} $$
$$ \text { unit vector } \stackrel{\rightharpoonup}{O C}=\frac{\stackrel{\rightharpoonup}{O C}}{|\stackrel{\rightharpoonup}{O C}|} $$
$$ \begin{aligned} & =\frac{-5 \underline{i}+12 \underline{j}}{\sqrt{(-5)^2+(12)^2}} \\ & =\frac{-5 \underline{i}+12 \underline{j}}{13} \\ & =\binom{-\frac{5}{13}}{\frac{12}{13}} \end{aligned} $$
Solutions by scale drawing are not accepted.
Diagram 5 shows two points on a Cartesian plane.
(a) $$
\text { Determine } \overrightarrow{A B} \text {. }
$$[1 mark]
(b) $$ \text { Point } C \text { lies on the same Cartesian plane. } $$
$$ \text { It is given that } \overrightarrow{O C}=(m+2) \underline{i}+12 \underline{j} \text {, such that } m \text { is a constant. } $$
(i) $$ \text { If } \overleftrightarrow{O C} \text { is parallel to } \overrightarrow{A B} \text {, } $$
$$ \text { find the value of } m \text { by using vector’s arithmetic operations. } $$
(ii) $$ \text { Hence, determine the unit vector in the direction of } \overrightarrow{O C} \text {. } $$
$$ \text { State your answer in the form of column vector. } $$
[5 marks]
Answer:
(a)(i) $$ \begin{aligned} & \overrightarrow{A B}=\overrightarrow{A O}+\overrightarrow{O B} \\ & \overrightarrow{A B}=-\overrightarrow{O A}+\overrightarrow{O B} \\ & \overrightarrow{A B}=-(6 \underline{i}-11 \underline{j})+(-4 \underline{i}+13 \underline{j}) \\ & \overrightarrow{A B}=-6 \underline{i}+11 \underline{j}-4 \underline{i}+13 \underline{j} \\ & \overrightarrow{A B}=-10 \underline{i}+24 \underline{j} \end{aligned} $$
(b)(i) $$ \overrightarrow{O C} \| \overrightarrow{A B} $$
$$ \begin{aligned} \overrightarrow{O C} & =k \overrightarrow{A B} \\ (m+2) \underline{i}+12 \underline{j} & =k(-10 \underline{i}+24 \underline{j}) \\ (m+2) \underline{i}+12 \underline{j} & =-10 k \underline{i}+24 k \underline{j} \end{aligned} $$
$$ \begin{aligned} j: 24 k & =12 \\ k & =\frac{12}{24} \\ k & =\frac{1}{2} \end{aligned} $$
$$ \begin{aligned} \underline{i}: m+2 & =-10 k \\ m+2 & =-10\left(\frac{1}{2}\right) \\ m & =-5-2 \\ m & =-7 \end{aligned} $$
$$ \therefore m=-7 $$
(b)(ii) $$ \begin{aligned} & \overrightarrow{O C}=(-7+2) \underline{i}+12 \underline{j} \\ & \overrightarrow{O C}=-5 \underline{i}+12 \underline{j} \end{aligned} $$
$$ \text { unit vector } \stackrel{\rightharpoonup}{O C}=\frac{\stackrel{\rightharpoonup}{O C}}{|\stackrel{\rightharpoonup}{O C}|} $$
$$ \begin{aligned} & =\frac{-5 \underline{i}+12 \underline{j}}{\sqrt{(-5)^2+(12)^2}} \\ & =\frac{-5 \underline{i}+12 \underline{j}}{13} \\ & =\binom{-\frac{5}{13}}{\frac{12}{13}} \end{aligned} $$