Questions: In Figure I, O A / / C B, O A perpendicular to O C, O A=2 and O B=1. It is given that cos angle A O B=2/3, vector OA= a and vector OB= b. (a) (i) Express vector CB in terms of a. (ii) Express vector OC in terms of a and b. (b) In Figure II, A D / / B O and O D perpendicular to D A. (i) Express vector OD in terms of a and b. (ii) P is a point on C D such that C P: P D=1: r. Express vector OP in terms of r, a and b. (iii) If M is a point on A C such that C M: M A=3: 1, find the ratio in which O M divides C D.

In Figure I, O A / / C B, O A perpendicular to O C, O A=2 and O B=1. It is given that cos angle A O B=2/3, vector OA= a and vector OB= b.
(a) (i) Express vector CB in terms of a.
(ii) Express vector OC in terms of a and b.
(b) In Figure II, A D / / B O and O D perpendicular to D A.
(i) Express vector OD in terms of a and b.
(ii) P is a point on C D such that C P: P D=1: r. Express vector OP in terms of r, a and b.
(iii) If M is a point on A C such that C M: M A=3: 1, find the ratio in which O M divides C D.

Transcript text: In Figure I, $O A / / C B, O A \perp O C, O A=2$ and $O B=1$. It is given that $\cos \angle A O B=\frac{2}{3}, \overrightarrow{O A}=\mathbf{a}$ and $\overrightarrow{O B}=\mathbf{b}$. (a) (i) Express $\overrightarrow{C B}$ in terms of $\mathbf{a}$. (ii) Express $\overrightarrow{O C}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. (b) In Figure II, $A D / / B O$ and $O D \perp D A$. (i) Express $\overrightarrow{O D}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. (ii) $P$ is a point on $C D$ such that $C P: P D=1: r$. Express $\overrightarrow{O P}$ in terms of $r, \mathbf{a}$ and $\mathbf{b}$. (iii) If $M$ is a point on $A C$ such that $C M: M A=3: 1$, find the ratio in which $O M$ divides $C D$.

Solution

Solution Steps

Step 1: Express $\overrightarrow{CB}$ in terms of $\mathbf{a}$

Given that $OA \parallel CB$ and $OA = 2$, $OB = 1$, and $\overrightarrow{OA} = \mathbf{a}$, we can deduce that $\overrightarrow{CB}$ is parallel to $\overrightarrow{OA}$ and has the same magnitude as $\overrightarrow{OA}$.

Since $OA = 2$ and $OB = 1$, the vector $\overrightarrow{CB}$ can be expressed as: \[ \overrightarrow{CB} = -\overrightarrow{OA} = -\mathbf{a} \]

Step 2: Express $\overrightarrow{OC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$

Given that $\cos \angle AOB = \frac{2}{3}$, $\overrightarrow{OA} = \mathbf{a}$, and $\overrightarrow{OB} = \mathbf{b}$, we need to find $\overrightarrow{OC}$.

Since $OA \perp OC$, $\overrightarrow{OC}$ is perpendicular to $\overrightarrow{OA}$. Using the given information, we can use the dot product to find the relationship between $\mathbf{a}$ and $\mathbf{b}$: \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \angle AOB = 2 \cdot 1 \cdot \frac{2}{3} = \frac{4}{3} \]

Since $OA \perp OC$, $\overrightarrow{OC}$ can be expressed as: \[ \overrightarrow{OC} = \mathbf{b} - \mathbf{a} \]

Step 3: Express $\overrightarrow{OD}$ in terms of $\mathbf{a}$ and $\mathbf{b}$

In Figure II, $AD \parallel BO$ and $OD \perp DA$. Given $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$, we need to find $\overrightarrow{OD}$.

Since $AD \parallel BO$, $\overrightarrow{AD}$ is parallel to $\overrightarrow{BO}$. Given that $OD \perp DA$, $\overrightarrow{OD}$ is perpendicular to $\overrightarrow{DA}$.

Using the given information, we can express $\overrightarrow{OD}$ as: \[ \overrightarrow{OD} = \mathbf{b} - \mathbf{a} \]