# Grade 10 Optional Mathematics: Vector Geometry
## Unit: Vector | Chapter: Geometric Proofs & Section Formula | Complete Solutions

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## 🎯 Key Concepts & Formulas

This section covers the fundamental principles of vector geometry, including addition laws, position vectors, and formulas for dividing a line segment. These concepts are the building blocks for the proofs and problems that follow.

### 1. Vector Addition

* **Triangle Law of Vector Addition:** If two vectors are represented by two sides of a triangle in sequence, then the third side of the triangle, taken in the opposite order, represents the sum (resultant) of the two vectors. For $\triangle ABC$, $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$.
* **Parallelogram Law of Vector Addition:** If two vectors acting simultaneously from a point are represented by the adjacent sides of a parallelogram, their resultant is represented by the diagonal of the parallelogram drawn from the same point. For parallelogram OABC with origin O, $\overrightarrow{OA} + \overrightarrow{OC} = \overrightarrow{OB}$.

### 2. Position Vector

The position vector of a point A, denoted as $\overrightarrow{OA}$ or $\overrightarrow{a}$, is the vector from a fixed origin (O) to that point. It's used to define the location of a point in space. The vector from point A to point B can be expressed using position vectors as $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$.

### 3. Section Formula

This formula finds the position vector of a point that divides the line segment joining two points in a specific ratio.

* **Internal Division:** If a point M divides the line segment joining P (position vector $\overrightarrow{p}$) and Q (position vector $\overrightarrow{q}$) internally in the ratio $m_1:m_2$, the position vector of M is:
    $$ \overrightarrow{OM} = \frac{m_1\overrightarrow{q} + m_2\overrightarrow{p}}{m_1 + m_2} $$
* **External Division:** If M divides the line joining A ($\overrightarrow{a}$) and B ($\overrightarrow{b}$) externally in the ratio $m:n$, its position vector is:
    $$ \overrightarrow{OM} = \frac{m\overrightarrow{b} - n\overrightarrow{a}}{m - n} $$

### 4. Mid-Point Formula

This is a special case of the internal section formula where the ratio is 1:1.

* **Formula:** The position vector of the mid-point M of the line segment joining points A and B is:
    $$ \overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} $$

### 5. Centroid Formula

The centroid is the point where the three medians of a triangle intersect.

* **Formula:** The position vector of the centroid G of a triangle ABC is the average of the position vectors of its vertices:
    $$ \overrightarrow{OG} = \frac{\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}}{3} $$

### 6. Scalar (Dot) Product and Perpendicularity

* **Condition for Perpendicularity:** Two non-zero vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular (orthogonal) if and only if their scalar product is zero.
    $$ \overrightarrow{a} \cdot \overrightarrow{b} = 0 $$

---

## 📐 Vector Proofs of Geometric Theorems

This section provides step-by-step proofs for key geometric theorems using vector methods.

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### 1. Angle in a Semi-circle is a Right Angle (Method 1)

**Prove by vector method that the angle at the circumference of a semi-circle is a right angle.**

**Given:**
* ACB is a semi-circle with center O.
* AB is the diameter.



**To Prove:** $\angle ACB = 90^\circ$

**Construction:** Join O to C.

**Proof:**

| Statement | Reason |
| :--- | :--- |
| i) $\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC}$ | By Triangle Law of Vector Addition. |
| ii) $\overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC}$ | By Triangle Law of Vector Addition. |
| iii) $\overrightarrow{BC} = -\overrightarrow{OA} + \overrightarrow{OC}$ | $\overrightarrow{BO} = -\overrightarrow{OA}$ since O is the center and OA and OB are radii in opposite directions. |
| iv) $\overrightarrow{AC} \cdot \overrightarrow{BC} = (\overrightarrow{OC} + \overrightarrow{OA}) \cdot (\overrightarrow{OC} - \overrightarrow{OA})$ | Taking the dot product of the rearranged vectors from (i) and (iii). |
| v) $\overrightarrow{AC} \cdot \overrightarrow{BC} = |\overrightarrow{OC}|^2 - |\overrightarrow{OA}|^2$ | Using the algebraic formula $(a+b)(a-b) = a^2 - b^2$. |
| vi) $\overrightarrow{AC} \cdot \overrightarrow{BC} = |\overrightarrow{OC}|^2 - |\overrightarrow{OC}|^2 = 0$ | $|\overrightarrow{OA}| = |\overrightarrow{OC}|$ (radii of the same circle). |
| vii) $\overrightarrow{AC} \perp \overrightarrow{BC}$ | The scalar (dot) product of two non-zero vectors is zero. |
| viii) $\angle ACB = 90^\circ$ | From statement (vii), the angle between the vectors is 90°. |

**Hence, the angle in a semi-circle is a right angle.**

***

### 2. Angle in a Semi-circle is a Right Angle (Method 2)

**PQ is a diameter of a circle with the centre X and M is a point on the circumference of the circle, prove that $\angle PMQ = 90^\circ$ by vector method.**

**Given:**
* PMQ is a semi-circle with center X.
* PQ is the diameter.



**To Prove:** $\angle PMQ = 90^\circ$

**Construction:** Join X to M.

**Proof:**

| Statement | Reason |
| :--- | :--- |
| i) $\overrightarrow{QM} = \overrightarrow{QX} + \overrightarrow{XM} = \overrightarrow{XM} - \overrightarrow{XQ}$ | By Triangle Law of Vector Addition. |
| ii) $\overrightarrow{PM} = \overrightarrow{PX} + \overrightarrow{XM} = \overrightarrow{XM} + \overrightarrow{PX}$ | By Triangle Law of Vector Addition. |
| iii) $\overrightarrow{PM} \cdot \overrightarrow{QM} = (\overrightarrow{XM} + \overrightarrow{PX}) \cdot (\overrightarrow{XM} - \overrightarrow{XQ})$ | Taking the dot product of (i) and (ii). |
| iv) $\overrightarrow{PM} \cdot \overrightarrow{QM} = (\overrightarrow{XM} + \overrightarrow{PX}) \cdot (\overrightarrow{XM} - \overrightarrow{PX})$ | Radii are equal, so $|\overrightarrow{XQ}| = |\overrightarrow{PX}|$. Since they are in opposite directions, $\overrightarrow{XQ} = -\overrightarrow{PX}$. |
| v) $\overrightarrow{PM} \cdot \overrightarrow{QM} = |\overrightarrow{XM}|^2 - |\overrightarrow{PX}|^2$ | Using the formula $(a+b)(a-b) = a^2-b^2$. |
| vi) $\overrightarrow{PM} \cdot \overrightarrow{QM} = |\overrightarrow{XM}|^2 - |\overrightarrow{XM}|^2 = 0$ | $|\overrightarrow{PX}| = |\overrightarrow{XM}|$ (radii of the same circle). |
| vii) $PM \perp QM$, so $\angle PMQ = 90^\circ$ | The scalar product of the vectors is zero. |

**Hence, the angle in a semi-circle is a right angle.**

***

<div id="isosceles"></div>
### 3. Median of an Isosceles Triangle

**Prove by vector method that the median from the vertex of an isosceles triangle is perpendicular to the base.**

**Given:**
* In $\triangle PQR$, $PQ = PR$.
* S is the mid-point of the base QR.



**To Prove:** $PS \perp QR$

**Proof:**

| Statement | Reason |
| :--- | :--- |
| i) $\overrightarrow{PS} = \frac{1}{2}(\overrightarrow{PQ} + \overrightarrow{PR})$ | S is the mid-point of QR, so $\overrightarrow{PS}$ is the median. |
| ii) $\overrightarrow{QR} = \overrightarrow{PR} - \overrightarrow{PQ}$ | By Triangle Law of Vector Addition ($\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR}$). |
| iii) $\overrightarrow{PS} \cdot \overrightarrow{QR} = \frac{1}{2}(\overrightarrow{PQ} + \overrightarrow{PR}) \cdot (\overrightarrow{PR} - \overrightarrow{PQ})$ | Taking the dot product of statements (i) and (ii). |
| iv) $\overrightarrow{PS} \cdot \overrightarrow{QR} = \frac{1}{2}(|\overrightarrow{PR}|^2 - |\overrightarrow{PQ}|^2)$ | Using the formula $(a+b)(a-b) = a^2-b^2$. |
| v) $\overrightarrow{PS} \cdot \overrightarrow{QR} = \frac{1}{2}(|\overrightarrow{PQ}|^2 - |\overrightarrow{PQ}|^2) = 0$ | Given $PQ=PR$, so $|\overrightarrow{PQ}| = |\overrightarrow{PR}|$. |
| vi) $PS \perp QR$ | Since the scalar product is zero. |

**Hence, the median to the base of an isosceles triangle is perpendicular to the base.**

***

<div id="rightAngled"></div>
### 4. Median to the Hypotenuse of a Right-Angled Triangle

**In a right-angled triangle, prove by vector method that the length of the median to the hypotenuse is half the length of the hypotenuse.**

**Given:**
* In right-angled triangle PQR, $\angle RPQ = 90^\circ$.
* S is the mid-point of the hypotenuse RQ.



**To Prove:** $PS = QS = RS$ (which implies $PS = \frac{1}{2}RQ$)

**Proof:**

| Statement | Reason |
| :--- | :--- |
| i) $\angle RPQ = 90^\circ$ | Given. |
| ii) $\overrightarrow{PR} \cdot \overrightarrow{PQ} = 0$ | The dot product of perpendicular vectors is zero. |
| iii) $\overrightarrow{PS} = \frac{1}{2}(\overrightarrow{PR} + \overrightarrow{PQ})$ | By the mid-point theorem, as S is the midpoint of RQ. |
| iv) $(\overrightarrow{PS})^2 = \frac{1}{4}(\overrightarrow{PR} + \overrightarrow{PQ})^2$ | Squaring both sides of statement (iii). |
| v) $PS^2 = \frac{1}{4}(|\overrightarrow{PR}|^2 + 2\overrightarrow{PR}\cdot\overrightarrow{PQ} + |\overrightarrow{PQ}|^2)$ | Expanding the square. |
| vi) $PS^2 = \frac{1}{4}(PR^2 + 0 + PQ^2)$ | From statement (ii), $\overrightarrow{PR} \cdot \overrightarrow{PQ} = 0$. |
| vii) $PS^2 = \frac{1}{4}(RQ^2)$ | In $\triangle PQR$, by Pythagorean theorem, $PR^2+PQ^2=RQ^2$. |
| viii) $PS = \frac{1}{2}RQ$ | Taking the square root of both sides. |
| ix) $PS = QS = RS$ | S is the mid-point of RQ, so $QS=RS=\frac{1}{2}RQ$. |

**Hence, the median to the hypotenuse is half the length of the hypotenuse.**

***

<div id="rhombus-perp"></div>
### 5. Diagonals of a Rhombus

**Prove by vector method that the diagonals of a rhombus are perpendicular.**

**Given:**
* PQRS is a rhombus where $|\overrightarrow{PQ}| = |\overrightarrow{QR}| = |\overrightarrow{RS}| = |\overrightarrow{SP}|$.



**To Prove:** $PR \perp QS$

**Proof:**

| Statement | Reason |
| :--- | :--- |
| i) $\overrightarrow{PR} = \overrightarrow{PQ} + \overrightarrow{PS}$ | By Parallelogram Law of Vector Addition. |
| ii) $\overrightarrow{QS} = \overrightarrow{PS} - \overrightarrow{PQ}$ | By Triangle Law of Vector Addition. |
| iii) $\overrightarrow{PR} \cdot \overrightarrow{QS} = (\overrightarrow{PS} + \overrightarrow{PQ}) \cdot (\overrightarrow{PS} - \overrightarrow{PQ})$ | Taking the dot product of (i) and (ii). |
| iv) $\overrightarrow{PR} \cdot \overrightarrow{QS} = |\overrightarrow{PS}|^2 - |\overrightarrow{PQ}|^2$ | Using the formula $(a+b)(a-b) = a^2-b^2$. |
| v) $\overrightarrow{PR} \cdot \overrightarrow{QS} = |\overrightarrow{PQ}|^2 - |\overrightarrow{PQ}|^2 = 0$ | All sides of a rhombus are equal, so $|\overrightarrow{PS}| = |\overrightarrow{PQ}|$. |
| vi) $PR \perp QS$ | Since the scalar product is zero. |

**Hence, the diagonals of a rhombus are perpendicular to each other.**

***

<div id='rectangle'></div>
### 6. Diagonals of a Rectangle

**Prove by vector method that the diagonals of a rectangle are equal.**

**Given:**
* PQRS is a rectangle.
* PR and QS are its diagonals.



**To Prove:** $PR = QS$

**Proof:**

| Statement | Reason |
| :--- | :--- |
| i) $\overrightarrow{PR} = \overrightarrow{PQ} + \overrightarrow{QR}$ | By Triangle Law of Vector Addition. |
| ii) $PR^2 = (\overrightarrow{PQ} + \overrightarrow{QR})^2 = |\overrightarrow{PQ}|^2 + 2\overrightarrow{PQ}\cdot\overrightarrow{QR} + |\overrightarrow{QR}|^2$ | Squaring both sides of (i). |
| iii) $PR^2 = PQ^2 + QR^2$ | Since adjacent sides of a rectangle are perpendicular, $\overrightarrow{PQ} \cdot \overrightarrow{QR} = 0$. |
| iv) $\overrightarrow{QS} = \overrightarrow{QR} + \overrightarrow{RS} = \overrightarrow{QR} - \overrightarrow{PQ}$ | By Triangle Law and since $\overrightarrow{RS} = -\overrightarrow{PQ}$. |
| v) $QS^2 = (\overrightarrow{QR} - \overrightarrow{PQ})^2 = |\overrightarrow{QR}|^2 - 2\overrightarrow{QR}\cdot\overrightarrow{PQ} + |\overrightarrow{PQ}|^2$ | Squaring both sides of (iv). |
| vi) $QS^2 = QR^2 + PQ^2$ | Since $\overrightarrow{QR} \cdot \overrightarrow{PQ} = 0$. |
| vii) $PR^2 = QS^2$ | From statements (iii) and (vi). |
| viii) $PR = QS$ | Taking the square root. |

**Hence, the diagonals of a rectangle are equal.**

***

### 7. Diagonals of a Parallelogram

**Prove by vector method that the diagonals of a parallelogram bisect each other.**

**Given:**
* OABC is a parallelogram.
* M and N are the midpoints of diagonals AC and OB respectively.



**To Prove:** The diagonals bisect each other (i.e., M and N are the same point, $\overrightarrow{OM} = \overrightarrow{ON}$).

**Proof:**

| Statement | Reason |
| :--- | :--- |
| i) $\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OC})$ | M is the midpoint of diagonal AC. |
| ii) $\overrightarrow{ON} = \frac{1}{2}(\overrightarrow{OB})$ | N is the midpoint of diagonal OB. |
| iii) $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC}$ | By Parallelogram Law of Vector Addition. |
| iv) $\overrightarrow{ON} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OC})$ | Substituting statement (iii) into (ii). |
| v) $\overrightarrow{OM} = \overrightarrow{ON}$ | From statements (i) and (iv). |

**Since the position vectors of the mid-points of both diagonals are the same, the mid-points coincide. Hence, the diagonals of a parallelogram bisect each other.**

***

### 8. Mid-Point Theorem for Quadrilaterals

**Prove by vector method that the quadrilateral formed by joining the mid-points of the sides of any quadrilateral is a parallelogram.**

**Given:**
* ABCD is a quadrilateral.
* P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively.



**To Prove:** PQRS is a parallelogram.

**Construction:** Join diagonal AC.

**Proof:**

| Statement | Reason |
| :--- | :--- |
| i) In $\triangle ABC$, $\overrightarrow{PQ} = \frac{1}{2}\overrightarrow{AC}$ | The line segment joining the mid-points of two sides of a triangle is parallel to and half the length of the third side. |
| ii) In $\triangle ADC$, $\overrightarrow{SR} = \frac{1}{2}\overrightarrow{AC}$ | Same reason as above. |
| iii) $\overrightarrow{PQ} = \overrightarrow{SR}$ | From statements (i) and (ii). This implies $PQ \parallel SR$ and $PQ = SR$. |
| iv) PQRS is a parallelogram. | A quadrilateral is a parallelogram if one pair of opposite sides is equal and parallel. |

**Hence, the quadrilateral formed by joining the mid-points is a parallelogram.**

***

### 9. Section Formula Proofs

#### a) Internal Division

**If the position vectors of points P and Q are $\overrightarrow{p}$ and $\overrightarrow{q}$ respectively, and point M divides the line segment PQ internally in the ratio $m_1:m_2$, prove that the position vector of M is $\overrightarrow{OM} = \frac{m_1\overrightarrow{q} + m_2\overrightarrow{p}}{m_1 + m_2}$.**



**Proof:**

| Statement | Reason |
| :--- | :--- |
| i) $\frac{PM}{MQ} = \frac{m_1}{m_2}$ | Given ratio. |
| ii) $m_2 \overrightarrow{PM} = m_1 \overrightarrow{MQ}$ | Cross-multiplication. Since P, M, Q are collinear and M is internal, the vectors are in the same direction. |
| iii) $m_2(\overrightarrow{OM} - \overrightarrow{OP}) = m_1(\overrightarrow{OQ} - \overrightarrow{OM})$ | By Triangle Law of Vector Addition. |
| iv) $m_2\overrightarrow{OM} - m_2\overrightarrow{p} = m_1\overrightarrow{q} - m_1\overrightarrow{OM}$ | Substituting given position vectors. |
| v) $m_1\overrightarrow{OM} + m_2\overrightarrow{OM} = m_1\overrightarrow{q} + m_2\overrightarrow{p}$ | Rearranging terms. |
| vi) $\overrightarrow{OM}(m_1 + m_2) = m_1\overrightarrow{q} + m_2\overrightarrow{p}$ | Taking $\overrightarrow{OM}$ common. |
| vii) $\overrightarrow{OM} = \frac{m_1\overrightarrow{q} + m_2\overrightarrow{p}}{m_1 + m_2}$ | Dividing by $(m_1+m_2)$. |

#### b) External Division

**If point M divides the line segment AB externally in the ratio $m:n$, prove that its position vector is $\overrightarrow{OM} = \frac{m\overrightarrow{b} - n\overrightarrow{a}}{m - n}$.**



**Proof:**

| Statement | Reason |
| :--- | :--- |
| i) $\frac{AM}{BM} = \frac{m}{n}$ | Given ratio. |
| ii) $n \overrightarrow{AM} = m \overrightarrow{BM}$ | Cross-multiplication. Vectors are in the same direction. |
| iii) $n(\overrightarrow{OM} - \overrightarrow{OA}) = m(\overrightarrow{OM} - \overrightarrow{OB})$ | By Triangle Law of Vector Addition. |
| iv) $n\overrightarrow{OM} - n\overrightarrow{a} = m\overrightarrow{OM} - m\overrightarrow{b}$ | Substituting position vectors $\overrightarrow{a}$ and $\overrightarrow{b}$. |
| v) $m\overrightarrow{b} - n\overrightarrow{a} = m\overrightarrow{OM} - n\overrightarrow{OM}$ | Rearranging terms. |
| vi) $m\overrightarrow{b} - n\overrightarrow{a} = \overrightarrow{OM}(m - n)$ | Taking $\overrightarrow{OM}$ common. |
| vii) $\overrightarrow{OM} = \frac{m\overrightarrow{b} - n\overrightarrow{a}}{m - n}$ | Dividing by $(m-n)$. |

---

## 🔢 Numerical Problems & Solutions

This section contains worked-out solutions for numerical problems applying the concepts of position vectors and section formulas.

***

### Problem 1: Finding the Centroid (from Median)
**In $\triangle ABC$, D is the midpoint of AC. If the position vectors of B and D are $3\overrightarrow{i} + 7\overrightarrow{j}$ and $3\overrightarrow{i} - 2\overrightarrow{j}$ respectively, find the position vector of the centroid G.**



**Solution:**
Let O be the origin.
Given: $\overrightarrow{OB} = 3\overrightarrow{i} + 7\overrightarrow{j}$ and $\overrightarrow{OD} = 3\overrightarrow{i} - 2\overrightarrow{j}$.
Since D is the mid-point of AC: $\overrightarrow{OD} = \frac{\overrightarrow{OA} + \overrightarrow{OC}}{2} \implies \overrightarrow{OA} + \overrightarrow{OC} = 2\overrightarrow{OD} = 6\overrightarrow{i} - 4\overrightarrow{j}$.
The position vector of the centroid G is:
$$
\begin{aligned}
\overrightarrow{OG} &= \frac{\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}}{3} = \frac{(\overrightarrow{OA} + \overrightarrow{OC}) + \overrightarrow{OB}}{3} \\
&= \frac{(6\overrightarrow{i} - 4\overrightarrow{j}) + (3\overrightarrow{i} + 7\overrightarrow{j})}{3} \\
&= \frac{9\overrightarrow{i} + 3\overrightarrow{j}}{3} = 3\overrightarrow{i} + \overrightarrow{j}
\end{aligned}
$$
**The position vector of G is $3\overrightarrow{i} + \overrightarrow{j}$.**

***

### Problem 2: Finding the Centroid (from Vertices)
**If the position vectors of the vertices A, B, and C of $\triangle ABC$ are $(3\overrightarrow{i} + 5\overrightarrow{j})$, $(5\overrightarrow{i} - \overrightarrow{j})$ and $(\overrightarrow{i} - 8\overrightarrow{j})$ respectively, find the position vector of the centroid G.**

**Solution:**
Given vertices:
* $\overrightarrow{OA} = 3\overrightarrow{i} + 5\overrightarrow{j}$
* $\overrightarrow{OB} = 5\overrightarrow{i} - \overrightarrow{j}$
* $\overrightarrow{OC} = \overrightarrow{i} - 8\overrightarrow{j}$

Using the centroid formula:
$$
\begin{aligned}
\overrightarrow{OG} &= \frac{\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}}{3} \\
&= \frac{(3\overrightarrow{i} + 5\overrightarrow{j}) + (5\overrightarrow{i} - \overrightarrow{j}) + (\overrightarrow{i} - 8\overrightarrow{j})}{3} \\
&= \frac{(3+5+1)\overrightarrow{i} + (5-1-8)\overrightarrow{j}}{3} \\
&= \frac{9\overrightarrow{i} - 4\overrightarrow{j}}{3} \\
&= 3\overrightarrow{i} - \frac{4}{3}\overrightarrow{j}
\end{aligned}
$$
**The position vector of G is $3\overrightarrow{i} - \frac{4}{3}\overrightarrow{j}$.**

***

### Problem 3: Finding a Mid-point Vector
**If the position vectors of two points A and B are $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\begin{pmatrix} 4 \\ 5 \end{pmatrix}$ respectively, find the position vector of K, the mid-point of AB.**

**Solution:**
Given: $\overrightarrow{OA} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}$.
$$
\begin{aligned}
\overrightarrow{OK} &= \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} \\
&= \frac{1}{2} \left[ \begin{pmatrix} 3 \\ 4 \end{pmatrix} + \begin{pmatrix} 4 \\ 5 \end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix} 7 \\ 9 \end{pmatrix} = \begin{pmatrix} 3.5 \\ 4.5 \end{pmatrix}
\end{aligned}
$$
**The position vector of K is $\begin{pmatrix} 3.5 \\ 4.5 \end{pmatrix}$.**

***

### Problem 4: Finding the Position Vector of P where MP = PN
**The position vectors of M and N are $9\overrightarrow{i} + 3\overrightarrow{j}$ and $3\overrightarrow{i} + 5\overrightarrow{j}$ respectively. Find the position vector of a point P such that $\overrightarrow{MP} = \overrightarrow{PN}$.**

**Solution:**
The condition $\overrightarrow{MP} = \overrightarrow{PN}$ means that P is the mid-point of the line segment MN.
Given:
* $\overrightarrow{OM} = 9\overrightarrow{i} + 3\overrightarrow{j}$
* $\overrightarrow{ON} = 3\overrightarrow{i} + 5\overrightarrow{j}$

Using the mid-point formula for P:
$$
\begin{aligned}
\overrightarrow{OP} &= \frac{\overrightarrow{OM} + \overrightarrow{ON}}{2} \\
&= \frac{(9\overrightarrow{i} + 3\overrightarrow{j}) + (3\overrightarrow{i} + 5\overrightarrow{j})}{2} \\
&= \frac{12\overrightarrow{i} + 8\overrightarrow{j}}{2} \\
&= 6\overrightarrow{i} + 4\overrightarrow{j}
\end{aligned}
$$
**The position vector of P is $6\overrightarrow{i} + 4\overrightarrow{j}$.**

***

### Problem 5: Finding a Vector from a Collinear Relationship
**In the given figure, $\overrightarrow{OA} = \overrightarrow{a}$ and $\overrightarrow{OB} = \overrightarrow{b}$. If $\overrightarrow{AC} = 3\overrightarrow{AB}$, find $\overrightarrow{OC}$.**



**Solution:**
We need to find $\overrightarrow{OC}$. Using the triangle law:
$$ \overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} $$
We are given $\overrightarrow{AC} = 3\overrightarrow{AB}$. First, let's find $\overrightarrow{AB}$:
$$ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \overrightarrow{b} - \overrightarrow{a} $$
Now substitute this into the expression for $\overrightarrow{AC}$:
$$ \overrightarrow{AC} = 3(\overrightarrow{b} - \overrightarrow{a}) $$
Finally, find $\overrightarrow{OC}$:
$$
\begin{aligned}
\overrightarrow{OC} &= \overrightarrow{OA} + \overrightarrow{AC} \\
&= \overrightarrow{a} + 3(\overrightarrow{b} - \overrightarrow{a}) \\
&= \overrightarrow{a} + 3\overrightarrow{b} - 3\overrightarrow{a} \\
&= 3\overrightarrow{b} - 2\overrightarrow{a}
\end{aligned}
$$
**The position vector $\overrightarrow{OC}$ is $3\overrightarrow{b} - 2\overrightarrow{a}$.**