01.05.00 – මිශ්‍ර ගැටලු

1. (2020)
   1. \underline a හා \underline b යනු ඒකක දෛශික දෙකක් යැයි ගනිමු.
      O මූලයක් අනුබද්ධයෙන් A, B හා C ලක්ෂය තුනක පිහිටුම් දෛශික පිළිවෙලින් 12\underline a\;,\;18\underline b හා 10\underline a+3\underline b වේ.
      \underline a හා \underline b ඇසුරෙන් \footnotesize{\overrightarrow{AC}} හා \scriptsize{\overrightarrow{CB}} ප්‍රකාශ කරන්න.
      A, B හා C ඒක රේඛීය බව අපෝහනය කර, AC : CB සොයන්න.
      {\small OC\;=\;\sqrt{139}} බව දී ඇත. {\small A\widehat OB=\frac\pi3} බව පෙන්වන්න.
      
      \begin{array}{rcl}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\small{\overrightarrow{AC}}\;&=&\;\small{\overrightarrow{AO}}\;+\;\small{\overrightarrow{OC}}\\&=&\;\small{\overrightarrow{OC}}\;-\;\small{\overrightarrow{OA}}\;\;\text{---}{\large ➄}\\&=&\;10\underline a\;+\;3\underline b\;-\;12\underline a\\&=&\;-2\underline a\;+\;3\underline b\;\;\text{---}{\large ➄}\\&&\\\small{\overrightarrow{CB}}\;&=&\;\small{\overrightarrow{OB}}\;-\;\small{\overrightarrow{OC}}\;\;\text{---}{\large ➄}\\&=&\;18\underline b\;-\;\left(10\underline a\;+\;3\underline b\right)\\&=&\;-10\underline a\;+\;15\underline b\;\;\text{---}{\large ➄}\\&&\\\small{\overrightarrow{CB}}\;&=&\;5\;\small{\overrightarrow{AC}}\;\;\text{---}{\large ➄}\end{array}
      \therefore\;A,\;B\;\mathrm{හා}\;C\;\mathrm{ඒක}\;\mathrm{රේඛීය}\;\mathrm{වන}\;\mathrm{අතර},\;\;\text{---}{\large ➄}
      \begin{array}{rcl}{AC}\;:\;{CB}\;&=&\;1\;:\;5\;\;\text{---}{\large ➄}\\&&\\{OC}\;&=&\;\sqrt{139}\;\;\text{---}\text{\large ➄}\\\small{\overrightarrow{{OC}}}\cdot\small{\overrightarrow{{OC}}}\;&=&\;139\;\;\text{---}{\large ➄}\\\left(10\underline{ a}\;+\;3\underline{ b}\right)\cdot\left(10\underline{ a}+3\underline{ b}\right)\;&=&\;139\;\;\text{---}{\large ➄}\\100\left|\underline{\underline{ a}}\right|^2\;+\;60\underline{ a}\cdot\underline{ b}\;+\;9\left|\underline{ b}\right|^2\;&=&\;139\;\;\text{---}{\large ➄}\\60\underline{ a}\cdot\underline{ b}\;&=&\;30\;\;\\\underline{ a}\cdot\underline{ b}\;&=&\;\frac12\;\;\text{---}{\large ➄}\\\left|\underline{ a}\right|\left|\underline{ b}\right|\;{Cos}\; A\widehat{ O} B\;&=&\;\frac12\;\;\text{---}{\large ➄}\\\therefore\; A\widehat{ O} B\;&=&\;\frac{\pi}3\;\;\text{---}{\large ➄}\\\end{array}
      \\
2. \underline a හා \underline b යනු නිශ්ශූන්‍ය අසමාන්තර දෛශික දෙකක් වන අතර \;α\underline a+\beta\underline b=0\; වෙයි. මෙහි α හා β අදිශ වෙයි. α = β = 0 බව පෙන්වන්න.
   A, B හා C ලක්ෂය වල O මූලය අනුබද්ධයෙන් පිහිටුම් දෛශික පිළිවෙලින් \;5\underline i+6\underline j\;,\;3\underline i-\underline j\;,\;-4\underline i+2\underline j වෙයි. P , Q , R ලක්ෂයයන් පිළිවෙලින් AB , BC , CA පාද මත පිහිටා ඇත්තේ AP : PB = 3 : 1 , BQ : QC = 2 : 1 ද, CR : RA = 2 : 3 වන පරිදිය.
   AQ සහ BR හි ඡේදන ලක්ෂයයෙහි පිහිටුම් දෛශිකය \frac{55}{143}\;\underline i\;+\;\frac{363}{143}\;\underline j බව පෙන්වන්න.

\begin{array}{l}\alpha \underline a+\beta \underline b\;=\;0\\\underline a\;\text{හා}\;\underline b\;\text{දෛශික}\;\text{අසමාන්තර}\;\text{බැවින්},\end{array}

\begin{array}{l}\underline a\end{array}≠\begin{array}{l}\underline b\;\text{හා}\;\underline a\end{array}≠\begin{array}{l}(-\underline b)\text{වේ}.\end{array}

\begin{array}{l}\text{තවද},\end{array}

\begin{array}{l}\underline a\end{array}≠\begin{array}{l}0\end{array} \begin{array}{l}\text{හා}\;\underline b\end{array}≠\begin{array}{l}0\end{array} \begin{array}{l}\;\text{බව}\;\text{දී}\;\text{ඇත.}\end{array}

\begin{array}{l}\therefore\;\alpha\underline a\;+\;\beta \underline b\;=\;0\;\text{වීමට}.\\\alpha=0\;\text{හා}\;\beta=0\;\text{විය}\;\text{යුතුය}.\;\\\therefore\;\alpha\;=\;\beta\;=\;0\;\text{වේ}\end{array}
\begin{array}{rcl}&&\\\;\;\;\;\;\;\;\;\;\;\;\;\overrightarrow{AB}\;&=&\;\overrightarrow{AO}\;+\;\overrightarrow{OB}\\&=&\;-5\;\underline i\;-6\;\underline j\;+\;3\;\underline i\;-\;\underline j\\&=&\;-2\;\underline i\;-\;7\;\underline j\\\\\overrightarrow{BC}\;&=&\;\overrightarrow{BO}\;+\;\overrightarrow{OC}\\&=&\;-3\;\underline i\;+\;\underline j\;-\;4\;\underline i\;+\;2\;\underline j\\&=&\;-7\;\underline i\;+\;3\;\underline j\;\\\\\overrightarrow{AC}\;&=&\;\overrightarrow{AO}\;+\;\overrightarrow{OC}\\&=&\;-5\;\underline i\;-\;6\;\underline j\;-\;4\;\underline i\;+\;2\;\underline j\\&=&\;-9\;\underline i\;-\;4\;\underline j\end{array}

AQ සහ BR හි ඡේදන ලක්ෂයය S ද,
BS : SR = 1 : λ ලෙසද, QS : SA = 1 : μ ලෙසද ගනිමු.
\begin{array}{rcl}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\overrightarrow{AQ}\;&=&\;\overrightarrow{AB}\;+\;\overrightarrow{BQ}\;\;\\&=&\;\overrightarrow{AB}\;+\frac23\;\overrightarrow{BC}\\[4px]&=&\;-2\;\underline i\;-\;7\;\underline j\;+\;\frac23\;(-7\;\underline i\;+\;3\;\underline j)\\[4px]&=&\;\left(\frac{-20}3\right)\;\underline i\;-\;\frac{15}3\;\underline j\\[4px]&=&\;\left(\frac{-20}3\right)\;\underline i\;-\;5\;\underline i\\[4px]\overrightarrow{BR}\;&=&\;\overrightarrow{BC}\;+\;\overrightarrow{CR}\;\;\\[4px]&=&\;\overrightarrow{BC}\;+\frac25\;\overrightarrow{CA}\\[4px]&=&\;-7\;\underline i\;+\;3\;\underline j\;-\;\frac25\;(-9\;\underline i\;-\;4\;\underline j)\\[4px]&=&\;\left(\frac{-17}5\right)\;\underline i\;-\;\frac{23}5\;\underline j\end{array}

SRA\;\triangle\text{යෙන්},

\begin{array}{rcl}\overrightarrow{AS}\;+\;\overrightarrow{SR}\;&=&\;\overrightarrow{AR}\;\\\left(\frac\mu{1+\mu}\right)\;\overrightarrow{AQ}\;+\;\left(\frac\lambda{1+\lambda}\right)\;\overrightarrow{BR}\;&=&\;\frac35\;\overrightarrow{AC}\;\\\left(\frac\mu{1+\mu}\right)\;\left(\frac{-20}3\right)\;\underline i\;-\frac{15}3\;\underline j\;+\;\left(\frac\lambda{1+\lambda}\right)\;\left[\frac{-17}5\;\underline i\;+\frac{23}5\;\underline j\right]\;&=&\;\frac35\;(-9\;\underline i\;-\;4\;\underline j)\end{array}

\underline i හා \underline j අසමාන්තර දෛශික බැවින් ඒවායෙහි සංගුණක සමාන කිරීමෙන්,

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\underline i සලකා,

\left(\frac{-20}3\right)\left(\frac\mu{1+\mu}\right)\;-\;\frac{17}5\left(\frac\lambda{1+\lambda}\right)\;=\;-\frac{27}5\;\text{---}\left(1\right)

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\underline j සලකා,

\left(\frac{-15}3\right)\left(\frac\mu{1+\mu}\right)\;-\;\frac{23}5\left(\frac\lambda{1+\lambda}\right)\;=\;-\frac{12}5\;\text{---}\left(2\right)

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)\;\times\;23\;+\;(2)\;\times\;17\;\text{න්,}

\begin{array}{rcl}\left(\frac{-460}3\right)\left(\frac\mu{1+\mu}\right)\;-\;\left(\frac{255}3\right)\left(\frac\mu{1+\mu}\right)\;&=&\;165\\[4px]\left(\frac{-715}3\right)\left(\frac\mu{1+\mu}\right)\;&=&\;165\\[4px]\left(\frac\mu{1+\mu}\right)\;&=&\;\frac{99}{143}\\[4px]143\mu\;&=&\;99\;+\;99\mu\\\mu\;&=&\;\frac{99}{44}\\[12px]\overrightarrow{OS}\;&=&\;\overrightarrow{OA}\;+\;\overrightarrow{AS}\\&=&\;5\underline i\;+\;6\underline j\;+\frac{\frac{99}{143}}{1+{\frac{99}{44}}}\left[\left(\frac{-20}3\right)\;\underline i\;-\;5\underline i\right]\\&=&\;\frac{\left(715\underline i\;-\;660\underline i\right)}{143}\;+\;\frac{\left(858\underline j\;-\;495\underline j\right)}{143}\\\overrightarrow{OS}\;&=&\;\frac{55}{143}\underline i\;+\;\frac{363}{143}\underline j\end{array}
\\

3. ABCD චතුරස්‍රයෙහි \overrightarrow{DC}=\underline a\;,\;\overrightarrow{AD}=\underline b\;\text{සහ}\;\overrightarrow{AB}=2\underline a වේ. AE : EB = 3 : 1 වන E ලක්ෂයය AB මත පිහිටා ඇත්තේ AB ⊥ CE වන පරිදිය. F යනු AD හි මධ්‍ය ලක්ෂයය වේ.
   EF හා AC හි ඡේදන ලක්ෂයය G වේ.
   \overrightarrow{FG}=\lambda\overrightarrow{FE}\;\text{විට,}\;\overrightarrow{AG}=\frac{3}{2}\lambda\underline a\;+\;\frac{1}{2}\left(1-\lambda\right)\underline b බව පෙන්වන්න.
   තවද, AG : GC හා FG : GE අනුපාත සොයන්න.
   B\widehat AD=\cos^{-1}\left(\frac{\left|\underline a\right|}{2\left|\underline b\right|}\right) බවද පෙන්වන්න.

\begin{array}{rcl}\overrightarrow{FE}\;&=&\;\overrightarrow{FA}\;+\;\overrightarrow{AE}\\&=&\;\overrightarrow{FA}\;+\;\frac34\overrightarrow{AB}‍\\[4px]&=&\;-\frac12\underline b\;+\;\frac34\left(2\underline a\right)\\[4px]&=&-\frac12\underline b\;+\;\frac32\underline a\\[4px]\overrightarrow{FG}\;&=&\;\lambda\overrightarrow{FE}\;\text{බැවින්,}\\\overrightarrow{GE}\;&=&\;\left(1-\lambda\right)\overrightarrow{FE}\\[4px]&=&\;\frac12\left(1-\lambda\right)\left(3\underline a-\underline b\right)\\[4px]\overrightarrow{AG}\;&=&\;\overrightarrow{AE}\;+\;\overrightarrow{EG}‍\\[4px]\;&=&\frac32\underline a\;+\;\frac12\left(1-\lambda\right)\left(\underline b-3\underline a\right)\\[4px]\overrightarrow{AG}\;&=&\;\frac32\lambda\underline a\;+\;\frac12\left(1-\lambda\right)\underline b\;\rightarrow\;\left(1\right)\\[4px]\overrightarrow{AG}\;&=&\;\mu\;\overrightarrow{AC}\;\text{නම්},\\&=&\;\mu\;\left(\overrightarrow{AD}+\overrightarrow{DC}\right)\\&=&\;\mu\;\left(\underline b+\underline a\right)\;\rightarrow\;\left(2\right)\end{array}

• \underline a\;≠\;0\;,\;\underline b\;≠\;0\;\text{හා}\;\underline a\;,\;\underline b\;\text{අසමාන්තර බැවින්,}

\begin{array}{rcl}\left\langle\left\langle\underline a\right\rangle\right\rangle\;\;\;\;\;\;\;\;\frac{3\lambda}2-\mu\;&=&\;0\\\mu\;&=&\;\frac{3\lambda}2\;\;\longrightarrow\;(3)\\\left\langle\left\langle\underline b\right\rangle\right\rangle\;\;\;\;\frac12-\frac\lambda2-\mu\;&=&\;0\end{array}

(3)න් ආදේශය:-

\begin{array}{rcl}\;\;\;\;\;\;\;\;\;\frac12-\frac\lambda2-\frac{3\lambda}2\;&=&\;0\\\lambda\;&=&\;\frac14\\FG\::\;GE\;&=&\;1\;:\;3\end{array}

(3)න්,

\mu\;=\;\frac38

4. 1. \underline a හා \underline b යනු නිශ්ශූන්‍ය සමාන්තර නොවූ දෛශික දෙකක් නම් \;\alpha\underline a+\beta\underline b=0\; වීමට α = 0 හා β = 0 විය යුතු බව පෙන්වන්න.
   2. සරල රේඛාවක් මගින් ABC ත්‍රිකෝණයේ දික්කල BC පාදය D හිදී ද, AC හා AB පාද පිළිවෙලින් E හා F හි දී අභ්‍යන්තරව ද ඡේදනය කරයි. මෙහි \frac{BD}{CD}=p(>1)\;,\;\frac{CE}{EA}\;=\;q\;\text{හා}\;\;\frac{AF}{FB}=r වේ.
      
      1. \overrightarrow{EF}\;=\;\frac1{1+q}\overrightarrow{CA}\;+\;\frac r{1+r}\overrightarrow{AB}\;\text{හා}
      2. \overrightarrow{DF}\;=\;\frac p{p-1}\overrightarrow{CA}\;+\;\frac{pr+1}{\left(1+r\right)\left(p-1\right)}\overrightarrow{AB}\;\text{බව පෙන්වන්න.}
         ඒ නයින්, pqr=1 බව අපෝහනය කරන්න.

\begin{array}{l}\alpha \underline a+\beta \underline b\;=\;0\\\underline a\;\text{හා}\;\underline b\;\text{දෛශික}\;\text{අසමාන්තර}\;\text{බැවින්},\end{array}

\begin{array}{l}\underline a\end{array}≠\begin{array}{l}\underline b\;\text{හා}\;\underline a\end{array}≠\begin{array}{l}(-\underline b)\text{වේ}.\end{array}

\begin{array}{l}\text{තවද},\end{array}

\begin{array}{l}\underline a\end{array}≠\begin{array}{l}0\end{array} \begin{array}{l}\text{හා}\;\underline b\end{array}≠\begin{array}{l}0\end{array} \begin{array}{l}\;\text{බව}\;\text{දී}\;\text{ඇත.}\end{array}

\begin{array}{l}\therefore\;\alpha\underline a\;+\;\beta \underline b\;=\;0\;\text{වීමට}.\\\alpha=0\;\text{හා}\;\beta=0\;\text{විය}\;\text{යුතුය}.\;\\\therefore\;\alpha\;=\;\beta\;=\;0\;\text{වේ}\end{array}

\begin{array}{l}\text{ii. }\;\;\;A,\;B,\;C,\;D,\;E\;\text{හා}\;F\;\text{හි පිහිටුම් පිළිවෙලින්}\;\underline a,\underline b,\underline c,\underline d,\underline e\;\text{හා}\;\underline f\;\text{ලෙස ගනිමු.}\end{array}

\begin{array}{rcl}\overrightarrow{CA}\;&=&\;\overrightarrow{CO}\;+\;\overrightarrow{OA}\\&=&\;\overrightarrow{OA}\;-\;\overrightarrow{OC}\\&=&\;\underline a\;-\;\underline c\\\overrightarrow{AB}\;&=&\;\overrightarrow{AO}\;+\;\overrightarrow{OB}\\&=&\;\overrightarrow{OB}\;-\;\overrightarrow{OA}\\&=&\;\underline b\;-\;\underline a\end{array}

\begin{array}{rcl}1.\;\overrightarrow{OE}\;&=&\;\frac{q\underline a+\underline c}{1+q}\\\overrightarrow{OF}\;&=&\;\frac{r\underline b+\underline a}{1+r}\\\overrightarrow{EF}\;&=&\;\overrightarrow{EO}\;+\;\overrightarrow{OF}\\&=&\;\overrightarrow{OF}\;-\;\overrightarrow{OE}\\&=&\;\frac{r\underline b+\underline a}{1+r}\;-\;\frac{q\underline a+\underline c}{1+q}\\&=&\;\frac{r\underline b+\underline a+rq\underline b+q\underline a-q\underline a-\underline c-rq\underline a-\underline{rc}}{\left(1+r\right)\left(1+q\right)}\\&=&\;\frac{r\left(\underline b-\underline c\right)+\left(\underline a-\underline c\right)+rq\left(\underline b-\underline a\right)}{\left(1+r\right)\left(1+q\right)}\\&=&\;\frac{r\left(\underline b-\underline a+\underline a-\underline c\right)+\left(\underline a-\underline c\right)+rq\left(\underline b-\underline a\right)}{\left(1+r\right)\left(1+q\right)}\\&=&\;\frac{\left(1+r\right)\left(\underline a-\underline c\right)}{\left(1+r\right)\left(1+q\right)}\;+\;\frac{r\left(1+q\right)\left(\underline b-\underline a\right)}{\left(1+r\right)\left(1+q\right)}\\&=&\;\frac{\underline a-\underline c}{1+q}\;+\;\frac{r\left(\underline b-\underline a\right)}{1+r}\\\overrightarrow{EF}\;&=&\;\left(\frac1{1+q}\right)\overrightarrow{CA}\;+\;\left(\frac r{1+r}\right)\overrightarrow{AB}\end{array}

\begin{array}{rcl}2.\;\overrightarrow{OD}\;&=&\;\frac{p\underline c-\underline b}{p-1}\\\overrightarrow{DF}\;&=&\;\overrightarrow{DO}\;+\;\overrightarrow{OF}\\&=&\;\frac{\underline b-p\underline c}{p-1}\;+\;\frac{r\underline b+\underline a}{1+r}\\&=&\;\frac{\underline b-p\underline c+r\underline b-pr\underline c+pr\underline b+p\underline a-r\underline b-\underline a}{\left(1+r\right)\left(p-1\right)}\\&=&\;\frac{\left(\underline b-\underline a\right)+p\left(\underline a-\underline c\right)+pr\left(\underline b-\underline c\right)}{\left(1+r\right)\left(p-1\right)}\\&=&\;\frac{\left(\underline b-\underline a\right)+p\left(\underline a-\underline c\right)+pr\left(\underline b-\underline a+a-\underline c\right)}{\left(1+r\right)\left(p-1\right)}\\&=&\;\frac{p\left(1+r\right)\left(\underline a-\underline c\right)}{\left(1+r\right)\left(p-1\right)}\;+\;\frac{\left(1+pr\right)\left(\underline b-\underline a\right)}{\left(1+r\right)\left(p-1\right)}\\\overrightarrow{DF}\;&=&\;\;\left(\frac p{p-1}\right)\overrightarrow{CA}\;+\;\left(\frac{pr+1}{\left(1+r\right)\left(p-1\right)}\right)\overrightarrow{AB}\end{array}

\begin{array}{rcl}\overrightarrow{OE}\;&=&\;\frac{q\underline a+\underline c}{1+q}\\\left(1+q\right)\underline e\;&=&\;q\underline a+\underline c\;\;\longrightarrow\;(1)\\\overrightarrow{OF}\;&=&\;\frac{r\underline b+\underline a}{1+r}\\\left(1+r\right)\underline f\;&=&{\;r\underline b+\underline a\;\;\longrightarrow\;(2)}\end{array}

(2)\times\;q\;-\;(1)\Rightarrow

\begin{array}{rcl}(1+r)\;q\underline f\;-\;(1+q)\underline e\;&=&\;rq\underline b\;+\;q\underline a\;-\;q\underline a\;-\;\underline c\\(1+r)\;q\underline f\;-\;(1+q)\underline e\;&=&\;rq\underline b\;-\;\underline c\end{array}

\begin{array}{rcl}\left(1+r\right)q\underline f\;-\;\left(1+q\right)\underline e\;&=&\;rq\underline b\;+\;q\underline a\;-\;q\underline a\;-\;\underline c\\\left(1+r\right)q\underline f\;-\;\left(1+q\right)\underline e\;&=&\;rq\underline b\;-\;\underline c\end{array}

දෙපසම (rq-1)න් බෙදීමෙන්,

\begin{array}{rcc}\frac{\left(1+r\right)q\underline f\;-\;\left(1+q\right)\underline e}{\left(rq-1\right)}&=&\frac{rq\underline b\;-\;\underline c}{rq-1}\\\overrightarrow{OD}&=&\;\frac{rq\underline b\;-\;\underline c}{rq-1}\end{array}

\therefore D\text{ ලක්ෂයයෙන්}\;BC\;\text{රේඛාව},\;\frac{BD}{CD}\;=\;\frac1{rq}\;\text{අනුපාතයට බෙදයි.}\;

\begin{array}{rcl}\frac{BD}{CD}\;&=&\;p\left(>1\right)\;\text{නිසා,}\\p\;&=&\;\frac1{rq}\\\text{එවිට},\;\;\;\;\;\;\;\;\;\;\\pqr\;&=&\;\frac1{rq}qr\\pqr\;&=&\;1\end{array}

5. \underline a හා \underline b යනු සමාන්තර නොවන්නාවුත් දෛශික 2 ක් අදිශ ගුණිතය අර්ථ දක්වන්න.
   ABC ත්‍රිකෝණයේ \overrightarrow {CA}=\underline a ද, \overrightarrow {CB}=\underline b වේ. \left(\underline a-\underline b\right)\cdot\left(\underline a-\underline b\right) අදිශ ගුණිතය සැලකීමෙන් 2ab\;\cos\;C\;=\;a\cdot a\;+\;b\cdot b\;-\;c\cdot c බව පෙන්වන්න. මෙහි \left|\underline a\right|=a\;,\;\left|\underline b\right|=b\;,\;\left|\underline c\right|=a-b වෙයි.

\;\;\;\underline a හා \underline b දෛශික අතර කෝණය α නම් අතර අදිශ ගුණිතය පහත පරිදි වෙයි.
\;\;\;\underline a\cdot\underline b\;=\;\left|\underline a\right|\left|\underline b\right|\;\cos\;\alpha

\begin{array}{rcl}\overrightarrow{CA}\;&=&\;\underline a\\\overrightarrow{CB}\;&=&\;\underline b\;\\\overrightarrow{AB}\;&=&\;\overrightarrow{AC}\;+\;\overrightarrow{CB}\\&=&\;\underline b\;-\;\;\underline a\\[6px]\left(\underline a-\underline b\right)\cdot\left(\underline a-\underline b\right)\;&=&\;\left|\underline a-\underline b\right|\cdot\left|\underline a-\underline b\right|\;\cos\;0\\&=&\;c\cdot c\cdot1\\&=&\;c^2\;\;\rightarrow\left(1\right)\end{array}

\begin{array}{rcl}\text{තවද,}\\‍\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(\underline a-\underline b\right)\cdot\left(\underline a-\underline b\right)\;&=&\;\underline a\cdot\underline a\;-\;2\;\underline a\cdot\underline b\;+\;\underline b\cdot\underline b\\&=&\;a^2\;\cos\;0\;-\;2\;\underline a\cdot\underline b\;+\;b^2\;\cos\;0\\&=&\;a^2\;-\;2\;\underline a\cdot\underline b\;+\;b^2\;\;\;;\;\;\;\text{සදහා ප්‍රකාශ ඉහත දී ඇති බැවින්}\\&=&\;a^2\;-\;2\;a\cdot b\;\cos\;C\;+\;b^2\;\;\rightarrow\;\left(2\right)\end{array}

\begin{array}{rcl}\;\;\;\;\;\left(1\right)\;=\;\left(2\right)\;\text{බැවින්,}\\\;\;\;\;\;c^2\;&=&\;a^2\;-\;2\;a\cdot b\;\cos\;C\;+\;b^2\;\\2\;a\cdot b\;\cos\;C\;&=&\;a^2\;+\;b^2\;-\;c^2\\\cos\;C\;&=&\;\frac{a^2\;+\;b^2\;-\;c^2}{2\;a\cdot b}\end{array}

ඉදිරියේදී ප්‍රශ්න ඇතුලත් වන්නේ මෙතනටයි.