## Math 10 Review chapter 2: Scalar product of two vectors and applications

## 1. Summary of theory

### 1.1. Trigonometric value of an angle

For each angle \(\alpha(0^o\leq \alpha\leq 180^o)\), we define a point M on the semicircle such that \(\widehat{MOx}=\alpha\). Let’s say the point M(x;y). Then:

- The y coordinate of the point M is called the sine of the angle \(\alpha\), we denote it \(sin\alpha\)
- The x coordinate of the point M is called the cosine of the angle \(\alpha\), we will denote it \(cos\alpha\).
- The ratio \(\frac{y}{x}\) \((x\neq 0)\) is called tan of angle \(\alpha\), we denote by \(tan\alpha\)
- The ratio \(\frac{x}{y}\) \((y\neq 0)\) is called the cotangent of the angle \(\alpha\), we denote it \(cot\alpha\)

### 1.2. Define dot product of two vectors

– The dot product of two vectors \(\vec a\) and \(\vec b\) is a number (algebraic quantity), denoted by \(\vec a.\vec b\) and defined determined by the formula

\(\vec a.\vec b=|\vec a|.|\vec b|.cos\left ( \vec a,\vec b \right )\)

– Coordinate expression of dot product: Given two vectors \(\vec{a}(x;y);\vec{b}(x’;y’)\). Then:

- \(\vec{a}.\vec{b}=xx’+yy’\)
- \(|\vec{a}|=\sqrt{x^2+y^2}\)
- \(cos(\vec{a};\vec{b})=\frac{xx’+yy’}{\sqrt{x^2+y^2}.\sqrt{x’^2+y’^ 2}},\vec{a}\neq \vec{0};\vec{b}\neq \vec{0}\)
- \(\vec{a}\perp \vec{b}\Leftrightarrow xx’+yy’=0\)

### 1.3. Theorem of cosine in triangle

– In triangle ABC, call \(Ab=c;AC=b;BC=a\), we have:

- \(a^2=b^2+c^2-2bc.cosA\)
- \(b^2=a^2+c^2-2ac.cosB\)
- \(c^2=a^2+b^2-2ab.cosC\)

From there, we have the following corollary:

- \(cosA=\frac{b^2+c^2-a^2}{2bc}\)
- \(cosB=\frac{a^2+c^2-b^2}{2ac}\)
- \(cosC=\frac{a^2+b^2-c^2}{2ab}\)

### 1.4. The sine theorem

- \(a=2RsinA, b=2RsinB, c=2RsinC\)
- \(\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R\)

### 1.5. The formula for the median of a triangle

- \(m_{a}^{2}=\frac{b^2+c^2}{2}-\frac{a^2}{4}\)
- \(m_{b}^{2}=\frac{a^2+c^2}{2}-\frac{b^2}{4}\)
- \(m_{c}^{2}=\frac{a^2+b^2}{2}-\frac{c^2}{4}\)

### 1.6. Formula to calculate the area of an extended triangle

- \(S=\frac{1}{2}a.h_a=\frac{1}{2}b.h_b=\frac{1}{2}c.h_c\)
- \(S=\frac{1}{2}ab.sinC=\frac{1}{2}ac.sinB=\frac{1}{2}bc.sinA\)
- \(S=\frac{abc}{4R}\)
- \(S=pr\)
- \(S=\sqrt{p(pa)(pb)(pc)}\)

## 2. Illustrated exercise

**Question 1: **Let \(ABC\) have \(AB=4\), \(AC=6\), \(\widehat{A}={{60}^{0}}\). Calculate the length of the side \(BC\) and the radius of the circumcircle of the triangle \(ABC\).

**Solution guide**

Applying the Cosic inequality we have

\(B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-2AB.AC.\cos A\)

\(={{4}^{2}}+{{6}^{2}}-2.4.6.\text{cos6}{{0}^{0}}=28\)

\(\Rightarrow BC=\sqrt{28}=2\sqrt{7}\)

We have

\(S=\frac{1}{2}AB.AC.\sin A\\\ = \frac{1}{2}.4.6.\sin {{60}^{0}}=6\sqrt{ 3}\\ S=\frac{abc}{4R}\)

\(\Rightarrow R=\frac{abc}{4S}=\frac{4.6.2\sqrt{7}}{4.6\sqrt{3}}=\frac{2\sqrt{21}}{3}\ )

**Verse 2: **In the plane \(Oxy\) gives the point \(A\left( 1;2 \right)\), \(B\left( 3;-4 \right)\). Let \(M\) be the midpoint of \(AB\).

Write the general equation of the line \(AB\). Calculate the distance from the point \(N\left( -2;1 \right)\) to the line \(AB\).

Write the general equation of the line \(d\) that passes through \(M\) and is perpendicular to the line \(\Delta :\,\,3x+y-5=0\).

**Solution guide**

a) \(\overrightarrow{AB}=\left( 2;-6 \right)\)

The line \(AB\) receives \(\overrightarrow{AB}=\left( 2;-6 \right)\) makes VTCP infer that the VTPT of \(AB\) is \(\overrightarrow{n}=\left ( 6,2 \right)\)

The line \(AB\) passes through \(A\left( 1;2 \right)\) and has a VTPT of \(\overrightarrow{n}=\left( 6;2 \right)\), so there is a method The generalization is \(6\left( x-1 \right)+2\left( y-2 \right)=0\)

\(\Leftrightarrow 6x+2y-10=0\)

\(d\left( N,AB \right)=\frac{\left| a{{x}_{0}}+b{{y}_{0}}+c \right|}{\sqrt{ {{a}^{2}}+{{b}^{2}}}}\)

\(=\frac{\left| 6.\left( -2 \right)+2.1-10 \right|}{\sqrt{{{6}^{2}}+{{2}^{2}} }}=\sqrt{10}\)

b) \(M\left( 2;-1 \right)\)

The VTPT of the line \(\Delta \) is \(\overrightarrow{{{n}_{\Delta }}}=\left( 3;1 \right)\)

\(d\) is perpendicular to \(\Delta \) so \(d\) gets the VTPT of \(\Delta \) as \(\overrightarrow{{{n}_{\Delta }}}=\left( 3;1 \right)\) as VTCP

Infer that the VTPT of \(d\) is \(\overrightarrow{n}=\left( 1;-3 \right)\).

\(d\) passes through \(M\left( 2;-1 \right)\) and has a VTPT of \(\overrightarrow{n}=\left( 1;-3 \right)\) so there is an equation in general \(1\left( x-2 \right)-3\left( y+1 \right)=0\\ \Leftrightarrow x-3y-5=0\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **In the plane \({\rm{Oxy}}\) gives the vector \(\vec a = ( – 5;2)\) and the vector \(\vec b = (3;-2)\) . Calculate the dot product \(\vec a.\vec b.\)

**Verse 2: **From the relation \({a^2} = {b^2} + {c^2} – 2bc.\cos A\) in the triangle, deduce the Pythagorean theorem.

**Question 3: **Prove that for any triangle ABC, we have \(a = 2R\sin A;b = 2R\sin B;c = 2R\sin C\), where \(R\) is the radius of the circumcircle triangle ABC.

### 3.2. Multiple choice exercises

**Question 1: **The triangle ABC has b=7, c=5 and \(cosA=\frac{3}{5}\). The area of triangle ABC is:

A. \(14\)

B. \(15\)

C. \(16\)

D. \(17\)

**Verse 2: **The radius of the circle inscribed in an equilateral triangle of side a is:

A. \(\frac{a\sqrt{3}}{4}\)

B. \(\frac{a\sqrt{3}}{5}\)

C. \(\frac{a\sqrt{3}}{6}\)

D. \(\frac{a\sqrt{3}}{7}\)

**Question 3: **Let ABC be a triangle with area S. If the length of side a is increased by 3 times, the length of side b is increased by 2 times and the magnitude of angle C is kept the same, the area of the new triangle created is:

A. 3S

B. 4S

C. 5S

D. 6S

**Question 4: **Let ABC be a triangle with BC=a, AC=b. Triangle ABC has the largest area when angle C is equal?

A. 60

B. 90

C. 150

D. 120

**Question 5: **In the coordinate plane Oxy, give 3 points \(A(-1;1),B(2;4),C(6;0)\). What is triangle ABC?

A. Pointed triangle

B. Right triangle

C. Obtuse triangle

D. Equilateral triangle

## 4. Conclusion

Through this lesson, you should know the following:

- Systematize all the knowledge learned in chapter Dot product of two vectors and applications
- Do the exercises of the chapter.

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