# Grade 10 Optional Mathematics: Scalar (Dot) Product
## Unit: Vector | Chapter: Scalar Product | Complete Solutions

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## 🎯 Chapter Introduction

This guide offers a complete overview of the **Scalar Product** (also known as the Dot Product) of vectors. It starts with a clear breakdown of the essential concepts and formulas, then provides detailed, step-by-step solutions to a wide range of problems, including finding the angle between vectors, determining vector magnitudes, and applying the condition for perpendicularity. 📐

### Key Learning Areas
1.  **Definition of Scalar Product:** Understanding what the scalar product is and how to calculate it.
2.  **Magnitude of a Vector:** Calculating the length or magnitude of a vector.
3.  **Angle Between Two Vectors:** Using the scalar product formula to find the angle between two vectors.
4.  **Condition for Perpendicularity:** Using the scalar product to prove that two vectors are perpendicular.
5.  **Properties of Unit Vectors:** Understanding the scalar products of unit vectors $\overrightarrow{i}$ and $\overrightarrow{j}$.

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## 🔑 Key Concepts & Formulas

Before diving into the problems, let's review the fundamental concepts.

### 1. The Scalar (Dot) Product

The scalar product of two vectors, $\overrightarrow{a}$ and $\overrightarrow{b}$, is a scalar (a number, not a vector) quantity.

* **Geometric Formula:**
    $$\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos\theta$$
    Where:
    * $|\overrightarrow{a}|$ is the magnitude (length) of vector $\overrightarrow{a}$.
    * $|\overrightarrow{b}|$ is the magnitude (length) of vector $\overrightarrow{b}$.
    * $\theta$ is the angle between the two vectors.



* **Component Formula:**
    If $\overrightarrow{a} = x_1\overrightarrow{i} + y_1\overrightarrow{j}$ and $\overrightarrow{b} = x_2\overrightarrow{i} + y_2\overrightarrow{j}$, then:
    $$\overrightarrow{a} \cdot \overrightarrow{b} = x_1x_2 + y_1y_2$$

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### 2. Magnitude of a Vector

The magnitude of a vector $\overrightarrow{a} = x\overrightarrow{i} + y\overrightarrow{j}$ is its length.

* **Formula:**
    $$|\overrightarrow{a}| = \sqrt{x^2 + y^2}$$

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### 3. Angle Between Two Vectors

We can rearrange the geometric formula to solve for the angle $\theta$.

* **Formula:**
    $$\cos\theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|}$$

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### 4. Condition for Perpendicular Vectors

Two non-zero vectors are perpendicular (orthogonal) if the angle between them is 90°. Since $\cos(90^\circ) = 0$, the condition is:

* **Condition:**
    $$\overrightarrow{a} \cdot \overrightarrow{b} = 0$$

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### 5. Properties of Unit Vectors $\overrightarrow{i}$ and $\overrightarrow{j}$

* The scalar product of a unit vector with itself is 1.
    $$\overrightarrow{i} \cdot \overrightarrow{i} = 1 \quad \text{and} \quad \overrightarrow{j} \cdot \overrightarrow{j} = 1$$
* The scalar product of two different standard unit vectors is 0 (as they are perpendicular).
    $$\overrightarrow{i} \cdot \overrightarrow{j} = 0 \quad \text{and} \quad \overrightarrow{j} \cdot \overrightarrow{i} = 0$$
* A vector dotted with itself equals the square of its magnitude:
    $$\overrightarrow{a} \cdot \overrightarrow{a} = |\overrightarrow{a}|^2$$

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## 📚 Exercise Solutions

### Short Questions

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#### Question 1: Finding Magnitude of a Vector
**If $\overrightarrow{p} \cdot \overrightarrow{q} = 6$, $|\overrightarrow{q}| = 2\sqrt{3}$ and the angle between $\overrightarrow{p}$ and $\overrightarrow{q}$ is 30°, find the length of $\overrightarrow{p}$.**

**Solution:**
Given: $\overrightarrow{p} \cdot \overrightarrow{q} = 6$, $|\overrightarrow{q}| = 2\sqrt{3}$, and $\theta = 30^\circ$.
We know the scalar product formula:
$$ \overrightarrow{p} \cdot \overrightarrow{q} = |\overrightarrow{p}| |\overrightarrow{q}| \cos\theta $$
Substituting the given values:
$$
\begin{aligned}
6 &= |\overrightarrow{p}| \cdot 2\sqrt{3} \cdot \cos(30^\circ) \\
6 &= |\overrightarrow{p}| \cdot 2\sqrt{3} \cdot \frac{\sqrt{3}}{2} \\
6 &= |\overrightarrow{p}| \cdot 3 \\
|\overrightarrow{p}| &= \frac{6}{3} = 2
\end{aligned}
$$
The length of $\overrightarrow{p}$ is **2 units**.

***

#### Question 2: Finding the Angle Between Two Vectors
**If $|\overrightarrow{a}| = 4$, $|\overrightarrow{b}| = 5$ and $\overrightarrow{a} \cdot \overrightarrow{b} = 10$, then find the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.**

**Solution:**
Using the formula for the angle between two vectors:
$$
\begin{aligned}
\cos\theta &= \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|} \\
\cos\theta &= \frac{10}{4 \cdot 5} \\
\cos\theta &= \frac{10}{20} = \frac{1}{2}
\end{aligned}
$$
Since $\cos(60^\circ) = \frac{1}{2}$, the angle is **$\theta = 60^\circ$**.

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#### Question 3: Finding the Scalar Product
**If $|\overrightarrow{a}| = 3\sqrt{3}$, $|\overrightarrow{b}| = 4$ and angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is 60°, then find the value of $\overrightarrow{a} \cdot \overrightarrow{b}$.**

**Solution:**
Using the scalar product formula:
$$
\begin{aligned}
\overrightarrow{a} \cdot \overrightarrow{b} &= |\overrightarrow{a}| |\overrightarrow{b}| \cos\theta \\
&= 3\sqrt{3} \cdot 4 \cdot \cos(60^\circ) \\
&= 12\sqrt{3} \cdot \frac{1}{2} \\
&= 6\sqrt{3}
\end{aligned}
$$
The value of $\overrightarrow{a} \cdot \overrightarrow{b}$ is **$6\sqrt{3}$**.

***

#### Question 4: Finding the Scalar Product
**If $|\overrightarrow{p}| = 6$, $|\overrightarrow{q}| = 6$ and the angle between them is 60°, find the value of $\overrightarrow{p} \cdot \overrightarrow{q}$.**

**Solution:**
Using the scalar product formula:
$$
\begin{aligned}
\overrightarrow{p} \cdot \overrightarrow{q} &= |\overrightarrow{p}| |\overrightarrow{q}| \cos\theta \\
&= 6 \cdot 6 \cdot \cos(60^\circ) \\
&= 36 \cdot \frac{1}{2} \\
&= 18
\end{aligned}
$$
The value is **18**.

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#### Question 5: Finding an Unknown in a Perpendicular Vector
**If the vectors $10\overrightarrow{i} - 7\overrightarrow{j}$ and $a\overrightarrow{i} + 10\overrightarrow{j}$ are perpendicular to each other, find the value of a.**

**Solution:**
Let $\overrightarrow{p} = 10\overrightarrow{i} - 7\overrightarrow{j}$ and $\overrightarrow{q} = a\overrightarrow{i} + 10\overrightarrow{j}$.
For two vectors to be perpendicular, their scalar product must be zero.
$$
\begin{aligned}
\overrightarrow{p} \cdot \overrightarrow{q} &= 0 \\
(10\overrightarrow{i} - 7\overrightarrow{j}) \cdot (a\overrightarrow{i} + 10\overrightarrow{j}) &= 0 \\
(10 \cdot a) + (-7 \cdot 10) &= 0 \\
10a - 70 &= 0 \\
10a &= 70 \\
a &= 7
\end{aligned}
$$
The value of **$a=7$**.

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#### Question 6: Finding an Unknown in a Perpendicular Vector
**If $\overrightarrow{a} = -5\overrightarrow{i} + 3\overrightarrow{j}$ and $\overrightarrow{b} = p\overrightarrow{i} + (p+2)\overrightarrow{j}$ are perpendicular to each other, find the value of p.**

**Solution:**
For perpendicular vectors, $\overrightarrow{a} \cdot \overrightarrow{b} = 0$.
$$
\begin{aligned}
(-5\overrightarrow{i} + 3\overrightarrow{j}) \cdot (p\overrightarrow{i} + (p+2)\overrightarrow{j}) &= 0 \\
(-5 \cdot p) + (3 \cdot (p+2)) &= 0 \\
-5p + 3p + 6 &= 0 \\
-2p &= -6 \\
p &= 3
\end{aligned}
$$
The value of **$p=3$**.

***

#### Question 7: Finding Magnitude of a Vector
**If $|\overrightarrow{b}| = 6$, $\overrightarrow{a} \cdot \overrightarrow{b} = 12$ and the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is 60°, find the value of $|\overrightarrow{a}|$.**

**Solution:**
Using the scalar product formula:
$$
\begin{aligned}
\overrightarrow{a} \cdot \overrightarrow{b} &= |\overrightarrow{a}| |\overrightarrow{b}| \cos\theta \\
12 &= |\overrightarrow{a}| \cdot 6 \cdot \cos(60^\circ) \\
12 &= |\overrightarrow{a}| \cdot 6 \cdot \frac{1}{2} \\
12 &= |\overrightarrow{a}| \cdot 3 \\
|\overrightarrow{a}| &= 4
\end{aligned}
$$
The magnitude of $\overrightarrow{a}$ is **4 units**.

***

#### Question 8: Finding the Angle Between Two Vectors
**Find the angle between the unit vector $\overrightarrow{i}$ and the vector $\overrightarrow{a} = \sqrt{3}\overrightarrow{i} + \overrightarrow{j}$.**

**Solution:**
Let the first vector be $\overrightarrow{u} = \overrightarrow{i} = 1\overrightarrow{i} + 0\overrightarrow{j}$.
Let the second vector be $\overrightarrow{a} = \sqrt{3}\overrightarrow{i} + 1\overrightarrow{j}$.

**Step 1: Find the scalar product.**
$$ \overrightarrow{u} \cdot \overrightarrow{a} = (1)(\sqrt{3}) + (0)(1) = \sqrt{3} $$
**Step 2: Find the magnitudes.**
$$ |\overrightarrow{u}| = |\overrightarrow{i}| = 1 $$
$$ |\overrightarrow{a}| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2 $$
**Step 3: Use the angle formula.**
$$
\begin{aligned}
\cos\theta &= \frac{\overrightarrow{u} \cdot \overrightarrow{a}}{|\overrightarrow{u}| |\overrightarrow{a}|} = \frac{\sqrt{3}}{1 \cdot 2} = \frac{\sqrt{3}}{2}
\end{aligned}
$$
Since $\cos(30^\circ) = \frac{\sqrt{3}}{2}$, the angle is **$\theta = 30^\circ$**.

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### Long Questions

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#### Question 9: Angle Between Unit Vectors
**If $\overrightarrow{a} + 2\overrightarrow{b}$ and $5\overrightarrow{a} - 4\overrightarrow{b}$ are perpendicular to each other and $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors, find the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.**

**Solution:**
Since the two vectors are perpendicular, their scalar product is zero.
$$ (\overrightarrow{a} + 2\overrightarrow{b}) \cdot (5\overrightarrow{a} - 4\overrightarrow{b}) = 0 $$
Expanding the dot product:
$$
\begin{aligned}
5(\overrightarrow{a} \cdot \overrightarrow{a}) - 4(\overrightarrow{a} \cdot \overrightarrow{b}) + 10(\overrightarrow{b} \cdot \overrightarrow{a}) - 8(\overrightarrow{b} \cdot \overrightarrow{b}) &= 0 \\
5|\overrightarrow{a}|^2 + 6(\overrightarrow{a} \cdot \overrightarrow{b}) - 8|\overrightarrow{b}|^2 &= 0
\end{aligned}
$$
Since $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors, $|\overrightarrow{a}|=1$ and $|\overrightarrow{b}|=1$.
$$
\begin{aligned}
5(1)^2 + 6(\overrightarrow{a} \cdot \overrightarrow{b}) - 8(1)^2 &= 0 \\
5 + 6(\overrightarrow{a} \cdot \overrightarrow{b}) - 8 &= 0 \\
6(\overrightarrow{a} \cdot \overrightarrow{b}) &= 3 \\
\overrightarrow{a} \cdot \overrightarrow{b} &= \frac{1}{2}
\end{aligned}
$$
Now, use the angle formula:
$$
\cos\theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|} = \frac{1/2}{1 \cdot 1} = \frac{1}{2}
$$
Thus, the angle is **$\theta = 60^\circ$**.

***

#### Question 10: Finding the Angle Between Two Vectors
**If $\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r} = 0$, $|\overrightarrow{p}| = 6$, $|\overrightarrow{q}| = 7$ and $|\overrightarrow{r}| = \sqrt{127}$, find the angle between $\overrightarrow{p}$ and $\overrightarrow{q}$.**

**Solution:**
From the given condition:
$$ \overrightarrow{p} + \overrightarrow{q} = -\overrightarrow{r} $$
Squaring both sides (taking the dot product with itself):
$$
\begin{aligned}
(\overrightarrow{p} + \overrightarrow{q}) \cdot (\overrightarrow{p} + \overrightarrow{q}) &= (-\overrightarrow{r}) \cdot (-\overrightarrow{r}) \\
|\overrightarrow{p}|^2 + 2(\overrightarrow{p} \cdot \overrightarrow{q}) + |\overrightarrow{q}|^2 &= |\overrightarrow{r}|^2
\end{aligned}
$$
Substitute the given magnitudes:
$$
\begin{aligned}
6^2 + 2(\overrightarrow{p} \cdot \overrightarrow{q}) + 7^2 &= (\sqrt{127})^2 \\
36 + 2(\overrightarrow{p} \cdot \overrightarrow{q}) + 49 &= 127 \\
85 + 2(\overrightarrow{p} \cdot \overrightarrow{q}) &= 127 \\
2(\overrightarrow{p} \cdot \overrightarrow{q}) &= 42 \\
\overrightarrow{p} \cdot \overrightarrow{q} &= 21
\end{aligned}
$$
Now, find the angle $\theta$:
$$
\cos\theta = \frac{\overrightarrow{p} \cdot \overrightarrow{q}}{|\overrightarrow{p}| |\overrightarrow{q}|} = \frac{21}{6 \cdot 7} = \frac{21}{42} = \frac{1}{2}
$$
Thus, the angle is **$\theta = 60^\circ$**.