Principle of Mathematical Induction

Mathematical Induction ကို domino effect ဖြင့် တင်စားဖော်ပြလေ့ရှိသည်။အောက်ပါ ပုံကိုလေ့လာကြည့်ပါ။



ပထမ domino လဲမကျလျှင် ၎င်းနောက်မှ စီထားသော domino များလဲကျမည်မဟုတ်ပါ။ ပထမ domino လဲကျသည်။ ထို့နောက် စီထားသော domino ထဲမှ ကြိုက်ရာတစ်ခုလဲကျသည်ဟု ယူဆပြီး၊ ၎င်းနောက်ရှိ ကပ်လျက်ရှိ‌သော domino လဲကျကြောင်း သက်သေပြနိုင်လျှင် မည်သည့် domino မဆို လဲကျသည်ဟု သက်သေပြနိုင်ပါသည်။ ဤသည်မှာ Mathematical Induction ဆိုင်ရာ သက်သေပြမှုတွင် အသုံးပြုသော အယူအဆဖြစ်သည်။

Mathematical Induction သည် အပေါင်းကိန်းပြည့် ဆိုင်ရာ အဆိုများ မှန်ကန်ချက်များကို သက်သေပြရာတွင် များစွာ အသုံးဝင်သည်။

Mathematical Induction ကို အသုံးပြု၍ သက်သေပြရာတွင် အဓိက အဆင့်နှစ်ဆင့်ဖြင့် သက်သေပြရမည်။

• Step (1) - ပေးထားသော အဆိုပြုချက်တစ်ခုအတွက် ပထမဆုံးအကြိမ် [ယျေဘုယျအားဖြင့် (n=1)] အတွက် မှန်ကန်ကြောင်း သက်သေပြရမည်။
• Step (2) - ထိုအဆိုသည် အခြား မည့်သည့် $k$ အကြိမ်အတွက်မဆို မှန်ကန်သည်ဟု သတ်မှတ်လိုက်ပြီး ၎င်းနောက်ရှိ ကပ်လျက်ရှိ‌သော $k+1$ အကြိမ်အတွက် မှန်ကြောင်း သက်သေပြနိုင်လျှင် ပေးထားသော အဆိုသည် အမြဲမှန်ကန်သည်။

Principle of Mathematical Induction

Mathematical Induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number. The technique involves two steps to prove a statement, as stated below −

Step 1 (Base step) − It proves that a statement is true for the initial value.

Step 2 (Inductive step) − It proves that if the statement is true for the $k^{\text{th}}$ iteration (or number $k$), then it is also true for $(k+1)^{\text{th}}$ iteration ( or number $k+1$).

Terminology of Statements

• Definition: an explanation of the mathematical meaning of a word.
စကားလုံးတစ်လုံး၏ သင်္ချာဆိုင်ရာ အဓိပ္ပာယ်ဖွင့်ဆိုချက်ကို Definition ဟုခေါ်သည်။


• Axiom: a basic assumption about a mathematical situation (model) which requires no proof.
သက်သေပြရန်မလိုအပ်သော သင်္ချာဆိုင်ရာ ယူဆချက်ကို Axiom ဟုခေါ်သည်။


• Theorem: a very important true statement that is provable in terms of definitions and axioms.
Definition , Axiom တို့ကို အသုံးပြု၍ အမြဲမှန်ကန်ကြောင်း သက်သေပြနိုင်သော အရေးပါသည့် သင်္ချာဆိုင်ရာ အဆိုပြုချက် ဖြစ်သည်။


• Proposition: a statement of fact that is true and interesting in a given context.
ပေးထားသော အကြောင်းအရာသည် အမြဲမှန်ကန်သည်ဟု ဆိုနိုင်သော သင်္ချာဆိုင်ရာ အဆိုပြုချက် ဖြစ်သည်။


• Lemma: a true statement used in proving other true statements.
အခြားအဆိုပြုချက်တစ်ခု မှန်ကန်ကြောင်း သက်သေပြရန် ကိုးကားအသုံးပြုသည့် အမြဲမှန်သော အကြောင်းအရာ ဖြစ်သည်။


• Corollary: a true statement that is a simple deduction from a theorem or proposition.
Theorem နှင့် Proporsition တို့မှ ဆင်းသက်လာသည့် ဆင့်ပွားမှန်ကန်ချက် ဖြစ်သည်။


• Proof: the explanation of why a statement is true.
အကြောင်းအရာတစ်ခု မှန်ကန်ကြောင်း ရှင်းပြချက်ကို သက်သေပြချက်ဟု ခေါ်သည်။


• Conjecture: a statement believed to be true, but for which we have no proof.
သက်သေပြခြင်း မရှိပဲ မှန်ကန်သည်ဟု ယုံကြည်ထားသော အဆိုတစ်ခုကို Conjecture ဟုခေါ်သည်။

Evaluation Steps (Steps of Proof)

• Step 1: Let $P(n)$ be a result or statement formulated in terms of $n$ (given question).


• Step 2: Prove that $P(1)$ or $P(\text{initial value})$ is true.


• Step 3: Assume that $P(k)$ is true.


• Step 4: Using Step 3 , prove that $P(k+1)$ is true.


• Step 5: Thus $P(1)$ is true and $P(k+1)$ is true whenever $P(k)$ is true.


• Conclusion: Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all natural numbers $n$.

Example (1)

Use mathematical induction to prove that $1 + 3 + 5 + \ldots + (2n - 1) = n^2$ is true for every positive integer $n$.

Solution

Let $P(n) : 1 + 3 + 5 + \ldots + (2n - 1) = n^2$.

When $n=1$,

LHS $= 1$

RHS $= 1^2 = 1$

$\therefore\ P(1)$ is true.

Assume that $P(k)$ is true for $n=k$.

$\therefore\ P(k) : 1 + 3 + 5 + \ldots + (2k - 1) = k^2 $

When $n=k+1$,

$ \quad 1 + 3 + 5 + \ldots + (2k - 1)+ \left[2(k+1) - 1\right] $

$ = k^2 + (2k + 2 - 1) $

$ = k^2 + 2k + 1 $

$ = (k+1)^2 $

$\therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all natural numbers $n$.

Example (2)

Use mathematical induction to prove that $1 + 2 + 3 + \ldots + n = \dfrac{n(n+1)}{2}$ is true for every $n\in\mathbb{N}$.

Solution

Let $P(n) : 1 + 2 + 3 + \ldots + n = \dfrac{n(n+1)}{2}$

When $n=1$,

LHS $= 1$

RHS $= \dfrac{1(1+1)}{2} = 1$

$\therefore\ P(1)$ is true.

Assume that $P(k)$ is true for $n=k$.

$\therefore\ P(k) : 1 + 2 + 3 + \ldots + k = \dfrac{k(k+1)}{2}$

When $n=k+1$,

$\quad 1 + 2 + 3 + \ldots + k+ (k+1) $

$ = \dfrac{k(k+1)}{2} + (k+1) $

$ = \dfrac{k^2+3k+2}{2} $

$ = \dfrac{k^2+2k+1+ k + 1}{2} $

$ = \dfrac{(k+1)^2+(k + 1)}{2} $

$ = \dfrac{(k + 1) \left[(k+1) + 1\right]}{2} $

$ \therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all positive integers $n$.

Example (3)

Prove that for any positive integer number $n$ , $n^3 + 2 n$ is divisible by $3$ by using mathematical induction.

Solution

Let $P(n) :n^3 + 2 n$ is divisible by $3$

When $n=1$,

$1^3+2(1) = 3$

Since $3$ is divisible by $3$, $P(1)$ is true.

When $n=k,\ P(k) :k^3 + 2 k$ is divisible by $3$.

Assume that $P(k)$ is true.

Let $k^3 + 2 k = 3p$ where $p$ is a positive integer.

When $n=k+1$,

$\quad (k + 1)^3 + 2 (k + 1) $

$ = k^3+3k^2+3k+1+2k+2 $

$ = (k^3+2k)+3k^2+3k+3$

$ = 3p+3(k^2+k+1)$

$ = 3(p+k^2+k+1)$

Since $p$ and $k$ are positive integers, $3(p+k^2+k+1)$ is an integer which is a multiple of $3$.

Therefore, $3(p+k^2+k+1)$ is divisible by $3$.

$\therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all positive integers $n$.

Example (4)

Prove that $n ! > 2 n$ where n is a positive integer and $n\ge 4$.

Solution

Let $P(n) :n ! > 2 n$

When $n=4$,

$4! = 4\times3\times2\times1=24$

$2(4) = 8$

$\therefore\ 24 > 8\Rightarrow 4!>2(4)$ $\therefore\ P(4)$ is true.

Assume that $P(k)$ is true for $n=k$ where $k>4$.

$\therefore\ P(k) :k! > 2k$.

When $n=k+1$,

$(k+1)!= (k+1)k! $

Since $k>4,\ k!> 2$.

$ \therefore\ (k+1)k! > 2(k+1) $

$ \therefore\ (k+1)! > 2(k+1) $

$\therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all positive integers $n$.

Example (5)

For every positive integer $n$, prove that $7^n – 3^n$ is divisible by $4$.

Solution

Let $P(n) :7^n – 3^n$ is divisible by $4$.

When $n=1$,

$7^1 – 3^1=7-3 = 4$

Since $4$ is divisible by $4$, $P(1)$ is true.

When $n=k,\ P(k) :7^k – 3^k$ is divisible by $4$.

Assume that $P(k)$ is true.

Let $7^k – 3^k = 4p$ where $p$ is a positive integer. When $n=k+1$,

$\quad 7^{k+1} – 3^{k+1} $

$ = 7\times7^k – 3\times3^k $

$ = 7\times7^k -(7\times3^k) +(7\times3^k) – (3\times3^k) $

$ = 7(7^k -3^k) +3^k(7 – 3) $

$ = 7(4p) +4\times3^k$

$ = 4 (7p +3^k)$

Since $p$ and $k$ are positive integers, $4 (7p +3^k)$ is an integer which is a multiple of $7$.

Therefore, $4 (7p +3^k)$ is divisible by $7$.

$\therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all positive integers $n$.

Example (6)

Using mathematical induction, show that $9^{n}-8 n-1$ where $n \in\mathbb{N}$ is divisible by $64$.

Solution

Let $P(n) :9^{n}-8 n-1$ where $n \in\mathbb{N}$ is divisible by $64$.

When $n=1$,

$9^{1}-8 (1)-1 = 0$

Since $0$ is divisible by $64$, $P(1)$ is true.

When $n=k,\ P(k) :9^{k}-8 k-1$ is divisible by $64$.

Assume that $P(k)$ is true.

Let $9^{k}-8 k-1 = 64p$ where $p$ is a positive integer.

When $n=k+1$,

$\quad 9^{k+1}-8 (k+1)-1$

$ = 9\times 9^{k}-8k-9 $

$ = 9\times 9^{k}-(9\times 8k)+ (9\times 8k) -8k-9 $

$ = 9\times 9^{k}-(9\times 8k)-9 + (9\times 8k) -8k$

$ = 9(9^{k}-8k-1) + 64k$

$ = 9(64p) + 64k$

$ = 64 (9p +k)$

Since $p$ and $k$ are positive integers, $64 (9p +k)$ is an integer which is a multiple of $64$.

Therefore, $64 (9p +k)$ is divisible by $64$.

$\therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all positive integers $n$.

Exercises Prove the following by using the principle of mathematical induction for all $n\in\mathbb{N}$: $1+3+3^{2}+\ldots+3^{n-1}=\dfrac{\left(3^{n}-1\right)}{2}$. $1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\left(\dfrac{n(n+1)}{2}\right)^{2}$. $1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+\ldots+\dfrac{1}{(1+2+3+\ldots n)}=\dfrac{2n}{n+1}$ $1\cdot2 \cdot3+2\cdot3 \cdot4+\ldots+n(n+1)(n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$. $1\cdot3+2\cdot3^{2}+3\cdot3^{3}+\ldots+n \cdot3^{n}=\dfrac{(2 n-1) 3^{n+1}+3}{4}$. $1\cdot2+2\cdot3+3\cdot4+\ldots+n \cdot(n+1)=\left[\dfrac{n(n+1)(n+2)}{3}\right]$. $1\cdot3+3\cdot5+5\cdot7+\ldots+(2 n-1)(2 n+1)=\dfrac{n\left(4 n^{2}+6 n-1\right)}{3}$. $1\cdot2+2\cdot2^{2}+3\cdot2^{3}+\ldots+n \cdot 2^{n}=(n-1) 2^{n+1}+2$. $\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\ldots+\dfrac{1}{2^{n}}=1-\dfrac{1}{2^{n}}$. $\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\ldots+\dfrac{1}{(3 n-1)(3 n+2)}=\dfrac{n}{(6 n+4)}$ $\dfrac{1}{1\cdot2 \cdot3}+\dfrac{1}{2\cdot3 \cdot4}+\dfrac{1}{3\cdot4 \cdot5}+\ldots+\dfrac{1}{n(n+1)(n+2)}=\dfrac{n(n+3)}{4(n+1)(n+2)}$. $a+a r+a r^{2}+\ldots+a r^{n-1}=\dfrac{a\left(r^{n}-1\right)}{r-1}$. $\left(1+\dfrac{3}{1}\right)\left(1+\dfrac{5}{4}\right)\left(1+\dfrac{7}{9}\right) \ldots\left(1+\dfrac{(2 n+1)}{n^{2}}\right)=(n+1)^{2}$. $\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right) \ldots\left(1+\dfrac{1}{n}\right)=(n+1)$. $1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}=\dfrac{n(2 n-1)(2 n+1)}{3}$. $\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+\ldots+\dfrac{1}{(3 n-2)(3 n+1)}=\dfrac{n}{(3 n+1)}$. $\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\ldots+\dfrac{1}{(2 n+1)(2 n+3)}=\dfrac{n}{3(2 n+3)}$. $1+2+3+\ldots+n<\dfrac{1}{8}(2 n+1)^{2}$. $n(n+1)(n+5)$ is a multiple of $3$ . $10^{2 n-1}+1$ is divisible by $11$ . $x^{2 n}-y^{2 n}$ is divisible by $x+y$. $3^{2 n+2}-8 n-9$ is divisible by $8$ . $41^{n}-14^{n}$ is a multiple of $27$ . $(2 n+7)< (n+3)^{2}$.

Sample Question for 2020 Matriculation Examination

2020

MATRICULATION EXAMINATION

Sample Question Set (3)

MATHEMATICS                        Time allowed: 3hours
WRITE YOUR ANSWERS IN THE ANSWER BOOKLET.
SECTION (A)
(Answer ALL questions.) 

1 (a).     Let $ \displaystyle f:R\backslash \{\pm 2\}\to R$ be a function defined by $ \displaystyle f(x)=\frac{{3x}}{{{{x}^{2}}-4}}$.Find the positive value of $z$ such that $f(z) = 1$.

(3 marks)

  (b).     If the polynomial $x^3 - 3x^2 + ax - b$ is divided by $(x - 2 )$ and $(x + 2)$, the remainders are $21$ and $1$ respectively. Find the values of $a$ and $b$.

(3 marks)

2(a).     Find the middle term in the expansion of $(x^2 - 2y)^{10}$.

(3 marks)

  (b).     In a sequence if $u_1=1$ and $u_{n+1}=u_n+3(n+1)$, find $u_5$ .

(3 marks)

3(a).     If $ \displaystyle P=\left( {\begin{array}{*{20}{c}} x & {-4} \\ {8-y} & {-9} \end{array}} \right)$ and $ \displaystyle {{P}^{{-1}}}=\left( {\begin{array}{*{20}{c}} {-3x} & 4 \\ {-7y} & 3 \end{array}} \right)$, find the values of $x$ and $y$.

(3 marks)

  (b).     A bag contains tickets, numbered $11, 12, 13, ...., 30$. A ticket is taken out from the bag at random. Find the probability that the number on the drawn ticket is

(i) a multiple of $7$

(ii) greater than $15$ and a multiple of $5$.

(3 marks)

4(a).     Draw a circle and a tangent $TAS$ meeting it at $A$. Draw a chord $AB$ making $ \displaystyle \angle TAB=\text{ }60{}^\circ $ and another chord $BC \parallel TS$. Prove that $\triangle ABC$ is equilateral.

(3 marks)

  (b).     If $ \displaystyle 3~\overrightarrow{{OA}}-2\overrightarrow{{OB}}-\overrightarrow{{OC}}~=\vec{0}$, show that the points $A, B$ and $C$ are collinear.

(3 marks)

5(a).     Solve the equation $2 \sin x \cos x -\cos x + 2\sin x - 1 = 0$ for $ \displaystyle 0{}^\circ \le x\le \text{ }360{}^\circ $.

(3 marks)

  (b).     Differentiate $ \displaystyle y=\frac{1}{{\sqrt[3]{x}}}$ from the first principles.

(3 marks)

SECTION (B)
(Answer Any FOUR questions.) 

6 (a).    Given that Given $ \displaystyle A=\{x\in R|\ x\ne -\frac{1}{2},x\ne \frac{3}{2}\}$. If $f:A\to A$ and $g:A\to A$ are defined by $f(x)=\displaystyle \frac{3x-5}{2x+1}$ and $g(x)=\displaystyle \frac{x+5}{3-2x}$, show that $f$ and $g$ are inverse of each other.

(5 marks)

   (b).     Given that $x^5 + ax^3 + bx^2 - 3 = (x^2 - 1) Q(x) - x - 2$, where $Q(x)$ is a polynomial. State the degree of $Q (x)$ and find the values of $a$ and $b$. Find also the remainder when $Q (x)$ is divided by $x + 2$.

(5 marks)

7 (a).     The binary operation $\odot$ on $R$ be defined by $x\odot y=x+y+10xy$ Show that the binary operation is commutative. Find the values $b$ such that $ (1\odot b)\odot b=485$.

(5 marks)

   (b).     If, in the expansion of $(1 + x)^m (1 – x)^n$, the coefficient of $x$ and $x^2$ are $-5$ and $7$ respectively, then find the value of $m$ and $n$.

(5 marks)

8 (a).     Find the solution set in $R$ for the inequation $2x (x + 2)\ge (x + 1) (x + 3)$ and illustrate it on the number line.

(5 marks)

   (b).     If the $ {{m}^{{\text{th}}}}$ term of an A.P. is $ \displaystyle \frac{1}{n}$ and $ {{n}^{{\text{th}}}}$ term is $ \displaystyle \frac{1}{m}$ where $m\ne n$, then show that $u_{mn} = 1$.

(5 marks)

9 (a).     The sum of the first two terms of a geometric progression is $12$ and the sum of the first four terms is $120$. Calculate the two possible values of the fourth term in the progression.

(5 marks)

   (b).     Given that $ A=\left( {\begin{array}{*{20}{c}} {\cos \theta } & {-\sin \theta } \\ {\sin \theta } & {\cos \theta } \end{array}} \right)$. If $A + A' = I$ where $I$ is a unit matrix of order $2$, find the value of $\theta$ for $ \displaystyle 0{}^\circ <\theta <90{}^\circ $.

(5 marks)

10 (a).   The matrix $A$ is given by $ \displaystyle A=\left( {\begin{array}{*{20}{c}} 2 & 3 \\ 4 & 5 \end{array}} \right)$.
(i) Prove that $A^2 = 7A + 2I$ where $I$ is the unit matrix of order $2$.
(ii) Hence, show that $ \displaystyle {{A}^{{-1}}}=\frac{1}{2}~\left( {A-7I} \right)$.

(5 marks)

   (b).   Draw a tree diagram to list all possible outcomes when four fair coins are tossed simultaneously. Hence determine the probability of getting:
(i) all heads,
(ii) two heads and two tails,
(iii) more tails than heads,
(iv) at least one tail,
(v) exactly one head.

(5 marks)

SECTION (C)
(Answer Any THREE questions.) 

11 (a).   $PQR$ is a triangle inscribed in a circle. The tangent at $P$ meet $RQ$ produced at $T$,and $PC$ bisecting $\angle RPQ$ meets side $RQ$ at $C$. Prove $\triangle TPC$ is isosceles.

(5 marks)

   (b).   In $\triangle ABC$, $D$ is a point of $AC$ such that $AD = 2CD$. $E$ is on $BC$ such that $DE \parallel AB$. Compare the areas of $\triangle CDE$ and $\triangle ABC$. If $\alpha (ABED) = 40$, what is $\alpha(ΔABC)$?

(5 marks)

12 (a).   If $L, M, N,$ are the middle points of the sides of the $\triangle ABC$, and $P$ is the foot of perpendicular from $A$ to $BC$. Prove that $L, N, P, M$ are concyclic.

(5 marks)

   (b).   Solve the equation $\displaystyle \sqrt{3}\cos \theta +\sin \theta =\sqrt{2}$ for $ \displaystyle 0{}^\circ \le \theta \le 360{}^\circ $.

(5 marks)

13 (a).   In $\triangle ABC, AB = x, BC = x + 2$, $AC = x – 2$ where $x > 4$, prove that $ \displaystyle \cos A=\frac{{x-8}}{{2(x-2)}}$. Find the integral values of $x$ for which $A$ is obtuse.

(5 marks)

  (b).   The sum of the perimeters of a circle and square is $k$, where $k$ is some constant. Using calculus, prove that the sum of their areas is least, when the side of the square is double the radius of the circle.

(5 marks)

14 (a).  The vector $ \overrightarrow{{OA}}$ has magnitude $39$ units and has the same direction as $ \displaystyle 5\hat{i}+12\hat{j}$. The vector $ \overrightarrow{{OB}}$ has magnitude $25$ units and has the same direction as $ \displaystyle -3\hat{i}+4\hat{j}$. Express $ \overrightarrow{{OA}}$ and $ \overrightarrow{{OB}}$ in terms of $ \hat{i}$ and $\hat{j}$ and find the magnitude of $ \overrightarrow{{AB}}.$

(5 marks)

   (b).  Find the coordinates of the stationary points of the curve $y = x\ln x - 2x$. Determine whether it is a maximum or a minimum point.

(5 marks)

Target Mathematics ၏ အစဉ်အလာအတိုင်း တက္ကသိုလ်ဝင်တန်း စာမေးပွဲကို ဝင်ရောက်ဖြေဆိုကြမည့် ကျောင်းသား/သူတို့အတွက် လေ့ကျင့်ရန် မေးခွန်းတစ်စုံ တင်ပြလိုက်ပါသည်။ လေ့ကျင့်ဖြေဆိုကြစေလိုပါသည်။ အဖြေများကို နောက်ရက်တွင် ဆက်လက် ဖော်ပြပေးပါမည်။

Position Vectors : Practice Problems

1.        Find the unit vector in the direction of $ \displaystyle \overrightarrow{{PQ}}$ where $ \displaystyle P$ and $ \displaystyle Q$ are points $ \displaystyle (2, 3)$ and $ \displaystyle (7, – 9)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ P\operatorname{and}\ Q\ \text{are points}\ (2,3)\ \operatorname{and}\ (7,-9).\\\\\therefore \ \ \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} 2 \\ 3 \end{array}} \right)\ \operatorname{and}\ \ \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} 7 \\ {-9} \end{array}} \right)\\\\\ \ \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\ \overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 7 \\ {-9} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 2 \\ 3 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 5 \\ {-12} \end{array}} \right)\\\\\therefore \ \ \left| {\ \overrightarrow{{PQ}}} \right|=\sqrt{{{{5}^{2}}+{{{\left( {-12} \right)}}^{2}}}}=13\\\\\therefore \ \ \text{The unit vector in }\\\ \ \ \ \text{the direction of}\ \ \ \ =\displaystyle \frac{{\overrightarrow{{PQ}}}}{{\left| {\ \overrightarrow{{PQ}}} \right|}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{13}}\left( {\begin{array}{*{20}{c}} 5 \\ {-12} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {\displaystyle \frac{5}{{13}}} \\ {-\displaystyle \frac{{12}}{{13}}} \end{array}} \right)\end{array}$

2.        Given that $ \displaystyle \overrightarrow{{OP}}=\widehat{\text{i}}+2\widehat{\text{j}}$ and $ \displaystyle \overrightarrow{{OQ}}=7\widehat{\text{i}}-4\widehat{\text{j}}$. Find the position vector of a point $ \displaystyle R$ which lies on the line $ \displaystyle PQ$ such that $ \displaystyle PR : RQ = 2 : 1$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OP}}=\widehat{\text{i}}+2\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OQ}}=7\widehat{\text{i}}-4\widehat{\text{j}}\\\\\ \ \ \ \ PR:RQ=2:1\\\\\ \ \ \ \ \text{By section formula},\ \\\\\ \ \ \ \ \ \overrightarrow{{OR}}=\displaystyle \frac{{\left( {1\times \overrightarrow{{OP}}} \right)+\left( {2\times \overrightarrow{{OQ}}} \right)}}{{1+2}}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{3}\ \left[ {\widehat{\text{i}}+2\widehat{\text{j}}+2\left( {7\widehat{\text{i}}-4\widehat{\text{j}}} \right)} \right]\\\\\ \ \ \ \ \ \ \ \ \ \ \ =5\widehat{\text{i}}-2\widehat{\text{j}}\end{array}$

3.        If $ \displaystyle \overrightarrow{{OA}}=-5\widehat{\text{i}}+6\widehat{\text{j}}$, $ \displaystyle \overrightarrow{{OB}}=2\widehat{\text{i}}+5\widehat{\text{j}}$ and $ \displaystyle \overrightarrow{{OC}}=9\widehat{\text{i}}+4\widehat{\text{j}}$, show that $ \displaystyle A, B$ and $ \displaystyle C$ are collinear.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=-5\widehat{\text{i}}+6\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OB}}=2\widehat{\text{i}}+5\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=9\widehat{\text{i}}+4\widehat{\text{j}}\\\ \ \ \ \ \\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {2\widehat{\text{i}}+5\widehat{\text{j}}} \right)-\left( {-5\widehat{\text{i}}+6\widehat{\text{j}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =7\widehat{\text{i}}-\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {9\widehat{\text{i}}+4\widehat{\text{j}}} \right)-\left( {2\widehat{\text{i}}+5\widehat{\text{j}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =7\widehat{\text{i}}-\widehat{\text{j}}\\\\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{BC}}\\\\\therefore \ \ \ A,B\ \operatorname{and}\ C\ \text{are collinear}.\end{array}$

4.        Given that $ \displaystyle \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} k \\ 5 \end{array}} \right)$, $ \displaystyle \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right)$ and $ \displaystyle \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 3 \\ {11} \end{array}} \right)$. If $ \displaystyle P, Q$ and $ \displaystyle R$ are collinear, find the value of $ \displaystyle k$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} k \\ 5 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 3 \\ {11} \end{array}} \right),\\\\\therefore \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} k \\ 5 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2-k} \\ 3 \end{array}} \right)\\\\\ \ \ \ \overrightarrow{{QR}}=\overrightarrow{{OR}}-\overrightarrow{{OQ}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 3 \\ {11} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right)\\\\\ \ \ \ \text{By the problem,}\ \\\\\ \ \ P,\ Q\ \operatorname{and}\ R\ \text{are collinear}.\\\\\therefore \ \ \text{Let}\ \overrightarrow{{PQ}}=h\overrightarrow{{QR}}\\\\\therefore \ \ \left( {\begin{array}{*{20}{c}} {-2-k} \\ 3 \end{array}} \right)=h\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right)\\\\\ \ \ \ \left( {\begin{array}{*{20}{c}} {-2-k} \\ 3 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {5h} \\ {3h} \end{array}} \right)\\\\\therefore \ \ 3h=3\\\\\ \ \ \ h=1\\\\\ \ \ \ -2-k=5h\\\\\ \ \ k=-2-5h\\\\\ \ \ k=-7\\\ \ \ \end{array}$

5.        Using a vector method, show that the points $ \displaystyle A (– 8, 10), B (– 1, 9)$ and $ \displaystyle C (6, 8)$ are collinear and hence find the ratio $ \displaystyle AB : BC$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=\left( {\begin{array}{*{20}{c}} {-8} \\ {10} \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OB}}=\left( {\begin{array}{*{20}{c}} {-1} \\ 9 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OC}}=\left( {\begin{array}{*{20}{c}} 6 \\ 8 \end{array}} \right),\\\\\therefore \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-1} \\ 9 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-8} \\ {10} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 7 \\ {-1} \end{array}} \right)\\\\\ \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 6 \\ 8 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-1} \\ 9 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 7 \\ {-1} \end{array}} \right)\\\\\therefore \ \ \overrightarrow{{AB}}=\overrightarrow{{BC}}\\\\\ \ \ A,\ B\ \operatorname{and}\ C\ \text{are collinear and }\\\\\ \ \ AB:BC=1:1\ \end{array}$

6.        If $ \displaystyle \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right)$, $ \displaystyle \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right)$ and $ \displaystyle \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right)$, show that $ \displaystyle \Delta PQR$ is a right triangle.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right),\\\\\therefore \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2} \\ 6 \end{array}} \right)\\\\\therefore \ PQ=\sqrt{{{{{\left( {-2} \right)}}^{2}}+{{6}^{2}}}}=\sqrt{{40}}\\\\\ \ \ \ \overrightarrow{{QR}}=\overrightarrow{{OR}}-\overrightarrow{{OQ}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {14} \\ {-2} \end{array}} \right)\\\\\therefore \ QR=\sqrt{{{{{14}}^{2}}+{{{\left( {-2} \right)}}^{2}}}}=\sqrt{{200}}\\\\\ \overrightarrow{{PR}}=\overrightarrow{{OR}}-\overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {12} \\ 4 \end{array}} \right)\\\\\therefore \ \ PR=\sqrt{{{{{12}}^{2}}+{{4}^{2}}}}=\sqrt{{160}}\\\\\ \ \ \ P{{Q}^{2}}+P{{R}^{2}}=40+160=200\\\\\ \ \ \ Q{{R}^{2}}=200\\\\\therefore \ \ P{{Q}^{2}}+P{{R}^{2}}=\ Q{{R}^{2}}\\\\\therefore \ \ \Delta PQR\ \text{is a right triangle}\text{.}\end{array}$

7.        If the position vectors of the points $ \displaystyle A, B$ and $ \displaystyle C$ are $ \displaystyle 9\widehat{\text{i}}+6\widehat{\text{j}}$, $ \displaystyle 4\widehat{\text{i}}+3\widehat{\text{j}}$ and $\displaystyle -5\widehat{\text{i}}+8\widehat{\text{j}}$ respectively, show that the $ \displaystyle \Delta ABC$ is an obtuse triangle.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=9\widehat{\text{i}}+6\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OB}}=4\widehat{\text{i}}+3\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=-5\widehat{\text{i}}+8\widehat{\text{j}}\\\\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =-5\widehat{\text{i}}-3\widehat{\text{j}}\\\\\therefore \ \ \ A{{B}^{2}}={{\left( {-5} \right)}^{2}}+{{\left( {-3} \right)}^{2}}=34\\\\\therefore \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =-9\widehat{\text{i}}+5\widehat{\text{j}}\\\\\therefore \ \ \ B{{C}^{2}}={{\left( {-9} \right)}^{2}}+{{5}^{2}}=106\\\\\therefore \ \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =-14\widehat{\text{i}}+2\widehat{\text{j}}\\\\\therefore \ \ \ A{{C}^{2}}={{\left( {-14} \right)}^{2}}+{{2}^{2}}=200\\\\\therefore \ \ \ A{{B}^{2}}+B{{C}^{2}}=140\\\\\therefore \ \ \ A{{C}^{2}}>A{{B}^{2}}+B{{C}^{2}}\\\\\therefore \ \ \ \Delta ABC\ \text{is an obtuse triangle}\text{.}\end{array}$

8.        The position vectors of the points A, B, and C are $ \displaystyle \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right)$, $ \displaystyle \left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right)$ and $ \displaystyle \left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)$ respectively. Prove that $ \displaystyle \Delta ABC$ is an isosceles right triangle.

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$ \displaystyle \begin{array}{l}\ \ \ \overrightarrow{{OA}}=\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right),\\\\\ \ \ \overrightarrow{{OB}}=\left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right),\\\\\ \ \ \overrightarrow{{OC}}=\left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)\\\\\ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2} \\ {-6} \end{array}} \right)\\\\\therefore \ \ AB=\sqrt{{{{{\left( {-2} \right)}}^{2}}+{{{\left( {-6} \right)}}^{2}}}}=\sqrt{{40}}\\\\\ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-4} \\ 8 \end{array}} \right)\\\\\therefore \ \ BC=\sqrt{{{{{\left( {-4} \right)}}^{2}}+{{8}^{2}}}}=\sqrt{{80}}\\\\\ \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-6} \\ 2 \end{array}} \right)\\\\\therefore \ \ AC=\sqrt{{{{{\left( {-6} \right)}}^{2}}+{{2}^{2}}}}=\sqrt{{40}}\\\\\therefore \ \ AB=AC\\\\\ \ \ \ A{{B}^{2}}+A{{C}^{2}}=40+40=80=B{{C}^{2}}\\\\\therefore \ \ \ \Delta ABC\ \text{is an isosceles right triangle}\text{.}\end{array}$

9.        $ \displaystyle OABC$ is a parallelogram such that $ \displaystyle \overrightarrow{{OA}}=5\widehat{\text{i}}+3\widehat{\text{j}}$ and $ \displaystyle \overrightarrow{{OC}}=-2\widehat{\text{i}}+\widehat{\text{j}}$. Find the unit vector in the direction of $ \displaystyle \ \overrightarrow{{OB}}$ and $ \displaystyle \ \overrightarrow{{AC}}$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=5\widehat{\text{i}}+3\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=-2\widehat{\text{i}}+\widehat{\text{j}}\\\\\ \ \ \ \ OABC\ \text{is a parallelogram}.\\\\\therefore \ \ \ \overrightarrow{{OB}}=\overrightarrow{{OA}}+\overrightarrow{{OC}}\ \ \ \left( {\because \text{parallelogram rule}\text{.}} \right)\\\\\ \ \ \ \ \overrightarrow{{OB}}=5\widehat{\text{i}}+3\widehat{\text{j}}-2\widehat{\text{i}}+\widehat{\text{j}}=3\widehat{\text{i}}+4\widehat{\text{j}}\\\\\therefore \ \ \ OB=\sqrt{{{{3}^{2}}+{{4}^{2}}}}=5\\\\\therefore \ \ \ \text{the unit vector in }\\\ \ \ \ \ \text{the direction of}\ \overrightarrow{{OB}}=\displaystyle \frac{{\overrightarrow{{OB}}}}{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{5}\left( {3\widehat{\text{i}}+4\widehat{\text{j}}} \right)\\\\\ \ \ \ \ \ \text{Again}\ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\ \ \\\ \ \ \ \ \overrightarrow{{AC}}=\left( {-2\widehat{\text{i}}+\widehat{\text{j}}} \right)-\left( {5\widehat{\text{i}}+3\widehat{\text{j}}} \right)=-7\widehat{\text{i}}-2\widehat{\text{j}}\\\\\therefore \ \ \ AC=\sqrt{{{{{\left( {-7} \right)}}^{2}}+{{{\left( {-2} \right)}}^{2}}}}=\sqrt{{53}}\\\\\therefore \ \ \ \text{the unit vector in }\\\ \ \ \ \ \text{the direction of}\ \overrightarrow{{AC}}=\displaystyle \frac{{\overrightarrow{{AC}}}}{{AC}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{\sqrt{{53}}}}\left( {-7\widehat{\text{i}}-2\widehat{\text{j}}} \right)\end{array}$

10.       Given that the position vectors of the points $ \displaystyle A, B$ and $ \displaystyle C$ relative to origin $ \displaystyle O$ are $ \displaystyle -\widehat{\text{i}}+\widehat{\text{j}}$, $ \displaystyle 5\widehat{\text{i}}+\widehat{\text{j}}$ and $ \displaystyle p\widehat{\text{i}}+q\widehat{\text{j}}$ respectively. If $ \displaystyle \Delta ABC$ is equilateral, find the possible values of $ \displaystyle p$ and $ \displaystyle q$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=-\widehat{\text{i}}+\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OB}}=5\widehat{\text{i}}+\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=p\widehat{\text{i}}+q\widehat{\text{j}}\\\\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =6\widehat{\text{i}}\\\\\therefore \ \ \ AB=6\\\\\ \ \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {p-5} \right)\widehat{\text{i}}+\left( {q-1} \right)\widehat{\text{j}}\\\\\ \ \ \ \ BC=\sqrt{{{{{\left( {p-5} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}\\\\\ \ \ \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ \ =\left( {p+1} \right)\widehat{\text{i}}+\left( {q-1} \right)\widehat{\text{j}}\\\\\ \ \ \ \ AC=\sqrt{{{{{\left( {p+1} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}\\\\\ \ \ \ \ \text{Since }\Delta ABC\ \text{is equilateral,}\\\\\ \ \ \ \ AB=BC=AC=6.\\\\\therefore \ \ \sqrt{{{{{\left( {p-5} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}=6\\\\\ \ \ \ {{\left( {p-5} \right)}^{2}}+{{\left( {q-1} \right)}^{2}}=36\\\\\ \ \ \ {{p}^{2}}+{{q}^{2}}-10p-2q=10\ ---(1)\\\\\ \ \ \ \text{Similarly,}\\\\\ \ \ \ \sqrt{{{{{\left( {p+1} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}=6\\\\\ \ \ \ {{\left( {p+1} \right)}^{2}}+{{\left( {q-1} \right)}^{2}}=36\\\\\ \ \ \ {{p}^{2}}+{{q}^{2}}+2p-2q=34\ ---(2)\\\\\ \ \ \ \text{By (2)}-\text{(1),}\\\text{ }\\\ \ \ \ 12p=24\\\\\therefore \ \ p=2\\\\\ \ \ \ \text{Substituting}\ p=2\text{ in (1),}\\\text{ }\\\therefore \ \ 4+{{q}^{2}}-20-2q=10\\\ \\\ \ \ \ {{q}^{2}}-2q=26\\\\\ \ \ \ {{q}^{2}}-2q+1=27\\\\\ \ \ \ {{(q-1)}^{2}}=27\\\\\ \ \ \ q-1=\pm \sqrt{{27}}\\\\\ \ \ \ q=1\pm 3\sqrt{3}\end{array}$

Trigonometry : Bearings and Directions

1.        Two ships leave a port at the same time. The first ship sails on a bearing of $ \displaystyle 032{}^\circ $ at $ \displaystyle 16$ km/h and the second on a bearing of $ \displaystyle 122{}^\circ $ at $ \displaystyle 24$ km/h. How far apart are they after $ \displaystyle 2.5$ hours?

2.       The bearing from point $ \displaystyle A$ to point $ \displaystyle B$ is $ \displaystyle S\ 55{}^\circ E$ and from a point $ \displaystyle B$ to point $ \displaystyle C$ is $ \displaystyle N\ 35{}^\circ E$. If a ship sails from $ \displaystyle A$ to $ \displaystyle B$, a distance of $ \displaystyle 81$ km, and from $ \displaystyle B$ to $ \displaystyle C$, a distance of $ \displaystyle 74$ km, how far is it from $ \displaystyle A$ to $ \displaystyle C$?

3.        A pilot flew $ \displaystyle 320$ miles at $ \displaystyle 068{}^\circ 1{0}'$, then $ \displaystyle 540$ miles at $ \displaystyle 158{}^\circ 1{0}'$. The pilot then flew back to his starting point. Find the length of the last flight.

4.        A ship leaves a harbour on a bearing of $ \displaystyle S\ 55{}^\circ E$ and travels for $ \displaystyle 120$ km. It then turns and continues on a bearing of $ \displaystyle N\ 35{}^\circ E$ for $ \displaystyle 62$ km. How far is the ship from the harbour.

5.        A boy could have walked along a straight road that heads $ \displaystyle N\ 30{}^\circ 3{7}' E$, but, instead, he walked due east $ \displaystyle 153.9$ yards and then due north to the end of the road. How far out of his way did the boy walk?

How far out of his way = (လမ်းအတိုင်းသွားရင် လျှောက်ရမယ့် အကွာအဝေးထက်) ဘယ်လောက် ပိုလျှောက်ခဲ့ရလဲ

6.        Two fishing boats start from a port. One travels $ \displaystyle 𝟏𝟓$ nautical miles per hour on a bearing of $ \displaystyle 025{}^\circ$ and the other travels $ \displaystyle 𝟏𝟖$ nautical miles per hour on a bearing of $ \displaystyle 𝟏𝟎𝟎°$. Assuming each maintains its course and speed, how far apart will the fishing boats be after two hours?

7.        A plane leaves from Yangon International Airport and travels due west at $ \displaystyle 570$ mi/hr. Another plane leaves $ \displaystyle 20$ minutes later and travels $ \displaystyle 22{}^\circ$ west of north at the rate of 585 mi/h. How far apart are they 40 minutes after the second plane leaves.

8.        An airplane travels on a bearing of $ \displaystyle 200{}^\circ$ for $ \displaystyle 𝟏𝟓𝟎𝟎$ miles and then changes to a bearing of $ \displaystyle 250{}^\circ$ and travels an additional $ \displaystyle 𝟓𝟎𝟎$ miles. How far is the airplane from its starting point?

9.        Two runners start from the same point. The first one heading due north at a constant speed of $ \displaystyle 12$ km/h. After $ \displaystyle 2.5$ hours, the second runner is at a bearing of $ \displaystyle 110{}^\circ$ from his starting point and 166.8 km away from the first.

(a) How far is the second runner from the starting point?

(b) How fast is the second runner runs on average?

10.       Yoon Yoon lives $ \displaystyle 𝟐$ miles west of her school and her friend May Kyaw lives $ \displaystyle 𝟑$ miles directly northeast of her.

(a) Draw a diagram representing this situation.

(b) How far does May Kyaw live from school? (c) What is the direction of school from May Kyaw's house?

11.       The tallest trees in the world are taller than a football field is long. Find the height of one of these trees, given the information in the figure.

12.       Two lighthouses are $ \displaystyle 𝟑𝟎$ miles apart on each side of shorelines running north and south, as shown. Each lighthouse keeper spots a boat in the distance. One lighthouse keeper notes the location of the boat as $ \displaystyle 40{}^\circ $ east of south, and the other lighthouse keeper marks the boat as $ \displaystyle 32{}^\circ $ west of south. What is the distance from the boat to each of the lighthouses at the time it was spotted?

13.       Two radar stations located $ \displaystyle 20$ miles apart both detect a UFO between them. The angle of elevation measured by the first station is $ \displaystyle 35$ degrees. The angle of elevation measured by the second station is $ \displaystyle 15$ degrees. What is the altitude of the UFO?

UFO = undefined flying object = ပန်းကန်ပြားပျံ
radar stations = ရေဒါစခန်းများ

14.       To measure the height of a hill, a surveyour measures the angle of elevation to the top of the hill to be $ \displaystyle 24$ degrees. He then moves back $ \displaystyle 200$ feet and measures the angle of elevation to be $ \displaystyle 22$ degrees. Find the height of the hill.

15.       A bridge is built across a small lake from a gazebo to a dock. The bearing from the gazebo to the dock is $ \displaystyle S\ 41{}^\circ W$. From a tree $ \displaystyle 100$ meters from the gazebo, the bearings to the gazebo and the dock are $ \displaystyle S\ 74{}^\circ E$ and $ \displaystyle S\ 28{}^\circ E$, respectively. Find the distance from the gazebo to the dock.

Gazebo = အပန်းဖြေစံအိမ်

16.       A boat is traveling due east parallel to the shoreline at a speed of $ \displaystyle 10$ miles per hour. At a given time, the bearing to a lighthouse is $ \displaystyle S\ 70{}^\circ E$, and $ \displaystyle 15$ minutes later the bearing is $ \displaystyle S\ 63{}^\circ E$. The lighthouse is located at the shoreline. What is the distance from the boat to the shoreline?

shoreline = ကမ်းပါး

17.       On a small lake, a boy swims from point of $ \displaystyle A$ to point of $ \displaystyle B$ at a bearing of $ \displaystyle N\ 28{}^\circ E$, then to point of $ \displaystyle C$ at a bearing of $ \displaystyle N\ 58{}^\circ W$, and finally back to point of $ \displaystyle A$, as shown in the figure below. Point of $ \displaystyle C$ lies of $ \displaystyle 800$ meters directly north of point of $ \displaystyle A$. Find the total distance that a boy swims.