01.02.00 – දෛශික 2 අතර ගුණිතය

01.අදිශ ගුණීතය(තිත් ගුණිතය)

අදිශ ගුණිතය යටතේ දෛශික 2ක් ගුණ කල විට ප්‍රතිඵලය ලෙස අදිශයක් ලැබේ.
මෙම ලැබෙන ප්‍රතිඵලය පහත පරිදි ප්‍රකාශ කල හැකිය.

a හා b යනු, 0 ලක්ශයක් අනුබද්දයෙන් A හා B ලක්ශ 2ක් එකිනෙකට (Φ) ආනත පිහිටුම් දෛශික 2ක් නම්,
\underline a\cdot\underline b\;=\;\left|\underline a\right|\left|\underline b\right|\cos\phi\; ලෙස අර්ත දැක්විය හැකිය.

අදිශ ගුණිතයේ ලක්ෂණ

1. Φ =90 ˚ විට,
   \begin{array}{rcl}\underline a\cdot\underline b\;&=&\;\left|\underline a\right|\left|\underline b\right|\;\cos\;\left(\phi\right)\\\underline a\cdot\underline b\;&=&\;\left|\underline a\right|\left|\underline b\right|\;\cos\;90^\circ\\\mathrm{නමුත්}\;\cos\;90^\circ\;&=&\;0\;\mathrm{බැවින්}\\\underline a\cdot\underline b\;&=&\;\left|\underline a\right|\left|\underline b\right|0\\\underline a\cdot\underline b\;&=&\;0\\\end{array}
   
   උදා:
   1. \\O\;\mathrm{ලක්ශයක්}\;\mathrm{අනුබද්දයෙන්}\;\mathrm{පිහිටි}\;\underline a\;\mathrm{හා}\;\underline b\;\mathrm{දෛශික}\;2\;\mathrm{ක්ද},\;\vert\underline a\vert=2\;\mathrm ද,\;\vert\underline b\vert=1\;\mathrm ද,\;\mathrm{නම්}\;\Phi\;=90\;\mathrm ද\;\mathrm{නම්},\;\underline a\cdot\underline b\;\;\;සොයන්න.\;
      \begin{array}{rcl}\underline a\cdot\underline b\;&=&\;\left|\underline a\right|\left|\underline b\right|\;\cos\;\left(\phi\right)\\\underline a\cdot\underline b\;&=&\;\left|\underline a\right|\left|\underline b\right|\;\cos\;90^\circ\\\mathrm{නමුත්}\;\cos\;90^\circ\;&=&\;0\;\mathrm{බැවින්}\\\underline a\cdot\underline b\;&=&\;0\\\end{array}
2. Φ =0 ˚ විට,
   \begin{array}{rcl}\underline a\;\cdot\underline b\;&=&\;\left|\underline a\right|\left|\underline b\right|cos\left(\phi\right)\\\underline a\;\cdot\underline b\;&=&\;\left|\underline a\right|\left|\underline b\right|cos\;0^\circ\\\mathrm{නමුත්}\;cos\;0\;˚\;&=&\;1\;\mathrm{බැවින්},\\\underline a\;\cdot\underline b\;&=&\;\left|\underline a\right|\left|\underline b\right|\\\end{array}
   උදා:
   1. \begin{array}{rcl}\\O\;\mathrm{ලක්ශයක්}\;\mathrm{අනුබද්දයෙන්}\;\mathrm{පිහිටි}\;\underline a\;\mathrm{හා}\;\underline b\;\mathrm{දෛශික}\;2\;\mathrm{අතර}\;\mathrm{කෝණය}\;O\;\mathrm ද,\vert\underline a\vert&=&4\;\mathrm ද,\vert\underline b\vert=3\;\mathrm ද\;\mathrm{නම්},\;\underline a\cdot\underline b\;\mathrm{සොයන්න}.\end{array}

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\underline a\;.\;\underline b\;=\;\left|\underline a\right|\left|\underline b\right|\;\cos\left(\Phi\right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\underline a\;.\;\underline b\;=\;\left|\underline a\right|\left|\underline b\right|\;\cos\;0^0\\\;\;\;\;\;\;\;\;\text{නමුත් }\;\;\cos\;0^0=1\;\text{බැවින්},\;\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;a\;.\;\underline b\;=\;4\;.\;3\;.\;1\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\underset{\;¯}a\;.\;\underline b\;=\;12

ත්‍රිමාණ අක්ෂ පද්ධතිය තුල ඒකක දෛශික අතර තිත් ගුණිතය

මෙහිදී පහත දැක්වෙන සම්බන්දතා භාවිතා කරයි.

• \begin{array}{rcl}\underline i\;\cdot\;\underline j\;&=&\;0\\\underline i\;\cdot\;\underline k\;&=&\;0\\\underline j\;\cdot\;\underline k\;&=&\;0\end{array}
• \begin{array}{rcl}\underline i\;\cdot\;\underline i\;&=&\;1\end{array}

අදිශ ගුණිතය න්‍යායදේශ න්‍යායට අනුකූල වේ.
එනම්,

\begin{array}{rcl}\underline a\;\cdot\;\underline b\;&=&\;\underline b\;\cdot\;\underline a\end{array} වේ.

අභ්‍යාස අංක 01

1. \begin{array}{rcl}\underline a\;\mathrm{හා}\;\underline b\;\mathrm{අතර}\;\mathrm{කෝණය}\;\mathrm{සොයන්න}.\end{array}
   \begin{array}{rcl}\;\;\;\mathrm i.\;\underline a&=&2\underline i+3\underline j\;\;\;\;\;\;\underline b\;&=&\;-2\underline i-2\underline j\\\mathrm{ii}.\;\underline a&=&5\underline i-\underline j\;\;\;\;\;\;\;\;\underline b\;&=&\;-2\underline i-2\underline j\\\mathrm{iii}.\;\underline a&=&\underline i+\underline j\;\;\;\;\;\;\;\;\;\underline b\;&=&\;\underline i-3\underline j\end{array}

\begin{array}{rcl}\;\;\;\mathrm i.\underline a&=&\;2\underline i+3\underline j\;\;\;\;\;\;\underline b\;=\;-2\underline i-2\underline j\end{array}\\\\\begin{array}{rcl}\;\underline{\;a}\;\cdot\;\underline b\;\;&=&\;\vert\underline a\vert\vert\underline b\vert\cos\;\phi\\(2\underline i+3\underline j)\cdot(-2\underline i-2\underline j)\;&=&\;\sqrt{2\cdot2+3\cdot3}\;\times\sqrt{2\cdot2+2\cdot2}\;\cos\;\phi\\-4\underline i\cdot\;\underline j-4\underline i\cdot\underline j\;-6\underline i\cdot\underline j\;-6\underline j\cdot\underline j\;\;&=&\;\sqrt{4+9}\times\sqrt{4+4}\;\cos\;\phi\\\;\;\;\;\;-4\times1\;-4\times0\;-6\times0\;-6\times1\;&=&\;\sqrt{13}\times\sqrt8\;\cos\;\phi\\-4\;-6\;&=&\;\sqrt{13}\times\sqrt8\;\cos\;\phi\\\cos\;\phi\;&=&\;\frac{-10}{\sqrt{13}\times2\sqrt2}\\\phi&=&\;\cos^{-1}\frac{-5}{\sqrt{13}\times\sqrt2}\end{array}

 

\begin{array}{rcl}\;\mathrm{ii}.\;\underline a&=&5\underline i-\underline j\;\;\;\;\;\;\;\underline b\;=\;-2\underline i-2\underline j\;\end{array}
\begin{array}{rcl}\;\underline{\;a}\;\cdot\;\underline b\;\;&=&\;\vert\underline a\vert\vert\underline b\vert\cos\;\phi\\(5\underline i-\underline j)\cdot(-2\underline i-2\underline j)\;&=&\;\sqrt{5\cdot5+1\cdot1}\;\times\sqrt{2\cdot2+2\cdot2}\;\cos\;\phi\\-10\;+2\;&=&\;\sqrt{26}\times\sqrt8\;\cos\;\phi\\-8\;&=&\;4\sqrt{13}\;\cos\;\phi\\-2\;&=&\;\sqrt{13}\;\cos\;\phi\\[4px]\cos\;\phi\;&=&\;\frac{-2}{\sqrt{13}}\\[4px]\phi&=&\;\cos^{-1}\frac{-2}{\sqrt{13}}\end{array}

\begin{array}{rc}\;\mathrm{iii}.\;\underline a&=\underline i\;+\;\underline j\;\;\;\;\;\;\;\underline b\end{array}=\;\underline i\;-\;3\underline j\\\\\begin{array}{rcl}\;\underline{\;a}\;\cdot\;\underline b\;\;&=&\;\vert\underline a\vert\vert\underline b\vert\cos\;\phi\\(\underline i+\underline j)\cdot(\underline i-3\underline j)\;&=&\;\sqrt{1\cdot1+1\cdot1}\;\times\sqrt{1\cdot1+3\cdot3}\;\cos\;\phi\\1-3\;&=&\;\sqrt2\times\sqrt{10}\;\cos\;\phi\\-2\;&=&\;2\sqrt5\;\cos\;\phi\\\cos\;\phi\;&=&\;\frac1{\sqrt5}\\\phi&=&\;\cos^{-1}\frac1{\sqrt5}\end{array}

2. \left|\underline a\right|=2\;ද,\;\left|\underline b\right|\;=3\;ද,\;වන\;අතර\;\underline{\mathrm a}\;\text{හා}\;\underline{\mathrm b}\;\text{අතර කෝණය}\;60^0\text{ ක් වේ.}\;\underline a\;.\;\underline b\;\text{සොයන්න.}\\\left(a+\;2b\right).\left(a-2b\right)\;\text{හා }\left|\mathrm a+2\mathrm b\right|\text{.}\left|a-2b\right|\text{ සොයන්න.එනයින්}\;a\;+\;2b\;\text{හා}\;a\;-\;2b\;\text{අතර කෝණය සොයන්න.}

02.දෛශික ගුණිතය(කතිර ගුණිතය)

\begin{array}{rcl}\underline a\cdot\underline b\;&=&\;\left|\underline a\right|\left|\underline b\right|\;Sin\;\phi\;\underline n\end{array}

n යනු, දෛශික දෙකේ කතිර ගුණිතයෙන් ලැබෙන දෛශිකයේ දිශාවට වූ ඒකක දෛශිකයවන අතර මෙය ෆ්ලෙමින්ගේ දකුණත් නියමයට අනුව ලබා ගත හැකිය.

කතිර ගුණිතයේ ලක්ශණ

1.\phi=0^\circ\\\begin{array}{rcl}\underline a\;\times\;\underline b\;&=&\;\vert\underline a\vert\vert\underline b\vert Sin\;\phi\;\underline n\;\\Sin\;0^\circ\;&=&\;0\;\;\mathrm{බැවින්},\;\\\underline a\;\times\;\underline b\;&=&\;0\;\mathrm{වේ}.\\&{}&{}\end{array}

2.\phi=90^\circ
\begin{array}{rcl}\underline a\;\times\;\underline b\;&=&\;\vert\underline a\vert\vert\underline b\vert Sin\;\phi\;\underline n\;\\Sin\;90^\circ\;&=&\;1\;\;\mathrm{බැවින්},\;\\\underline a\;\times\;\underline b\;&=&\;{\vert\underline a\vert\vert\underline b\vert\underline n}\;\mathrm{වේ}.\\\end{array}

\begin{array}{l}3.\;\;\;\underline a\;=\;\underline b\;\mathrm{නම්},\;\\\underline a\;x\;\underline b\;=\;\left|\underline a\right|\left|\underline b\right|\sin\Phi\;\mathrm n\\\;\;\;\;\;\;\;\underline a\;=\;\;\underline b\;\text{බැවින්,}\\\;\;\;\;\;\left|\underline a\right|\;=\left|\underline b\right|\;\text{ හා}\\\;\;\;\;\;\;\Phi\;\;=\;0\;\text{බැවින්,}\;\\\;\;\sin\;0^0=0\;\text{වේ.}\\\;\;\underline a\;x\;\underline b\;=\left|\underline a\right|.\left|\underline a\right|.0.\text{n}\;\\\therefore\;\underline a\;x\;\underline b\;=\;0\;\text{වේ.}\\\;\;\;\;\;\;\;\;\;\end{array}

ඒකක දෛශික අතර කතිර ගුණිතය

\begin{array}{rcl}\underline i\times\underline i\;&=&\;\left|\underline i\right|\left|\underline i\right|\;Sin\;0^\circ\;\underline n\\\underline i\times\underline i\;&=&\;0\\\underline j\times\underline j\;&=&\;0\\\underline k\times\underline k\;&=&\;0\end{array} \begin{array}{rcl}\underline i\times\underline j\;&=&\;\left|\underline i\right|\left|\underline j\right|\;Sin\;90^\circ\;\underline n\\\underline i\times\underline j\;&=&\;0\\\underline j\times\underline k\;&=&\;0\\\underline i\times\underline k\;&=&\;0\\\end{array}

දෛශික ගුණිතය න්‍යායදේශ වීම සිදු නොවේ.
එනම්,

\begin{array}{rcl}\underline a\times\underline b\;\end{array}≠\begin{array}{rcl}\;\underline b\;\times\;\underline a\;\mathrm{වේ}.\end{array}

අභ්‍යාස

1. \vert\underline a\vert=2 ද,\;\vert\underline b\vert=\;3\; ද,\;\underline a\;හා\;\underline b\;අතර කෝණය 120˚ ද, නම් \underline a\;\cdot\;\underline b\;සොයන්න.\;(\underline a+2\underline b)\cdot(\underline a-2\underline b)\;හා\;\;\vert\underline a+2\underline b\vert\cdot\vert\underline a-2b\vert\;ගණනය කර එනයින්\;(\underline a+2\underline b)\cdot(\underline a-2\underline b)\;ගණනය කරන්න.

\begin{array}{rcl}\underline a\cdot\underline b\;&=&\;\left|\underline a\right|\left|\underline b\right|\;\cos\;\phi\\\underline a\cdot\underline b\;&=&\;2\times3\times\left(\frac{-1}2\right)\\\underline a\cdot\underline b\;&=&\;-3\end{array} \begin{array}{rcl}\left(\underline a+2\underline b\right)\cdot\left(\underline a-2\underline b\right)\;&=&\;\underline a^2+2\underline a\cdot\underline b-2\underline a\cdot\underline b-4\underline b^2\\&=&\left|\underline a\right|^2-\left|4\underline b\right|^2\\&=&2^2-4\times3^2\\&=&4-36\\&=&-32\end{array} \begin{array}{rcl}\left|\underline a+2\underline b\right|\cdot\left|\underline a-2\underline b\right|\;&=&\;\sqrt{\left(\underline a+2\underline b\right)^2\;\times\;\left(\underline a-2\underline b\right)^2}\\&=&\;\sqrt{\left(\underline a^2+4\underline a\cdot\underline b+4\underline{b^2}\right)\;\times\;\left(\underline a^2-4\underline a\cdot\underline b+4\underline{b^2}\right)}\\&=&\;\sqrt{\left(\left|\underline a\right|^2+4\underline a\cdot\underline b+4\left|\underline b\right|^2\right)\;\times\;\left(\left|\underline a\right|^2-4\underline a\cdot\underline b+4\left|\underline b\right|^2\right)}\\&=&\;\sqrt{\left(2^2+4\left(-3\right)+4\times3^2\right)\;\times\;\left(2^2-4\underline a\cdot\underline b+4\times3^2\right)}\\&=&\;\sqrt{\left(4-12+36\right)\;\times\;\left(4+12+36\right)}\\&=&\sqrt{28\times52}\\&=&4\sqrt{91}\end{array} \begin{array}{rcl}(\underline a+2\underline b)\cdot(\underline a-2\underline b)\;&=&\;\vert\underline a+2\underline b\vert\cdot\vert\underline a-2\underline b\vert\;\cos\;\phi\\-32\;&=&\;4\sqrt{91}\;\cos\;\phi\\-8\;&=&\;\sqrt{91}\;\cos\;\phi\\\phi\;&=&\;\cos^{-1}\left(\frac{-8}{\sqrt{91}}\right)\end{array}

ඉදිරියේදී ප්‍රශ්න ඇතුලත් වන්නේ මෙතනටයි.