11.3-bo'lim: Skalyar ko'paytma
Navbatdagi mavzumiz — vektorlarning skalyar ko'paytmasi. To'g'ridan-to'g'ri skalyar ko'paytmaning ta'rifiga o'tamiz. Berilgan ikki vektor \(\vec a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \) va \(\vec b = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \) uchun ularning skalyar ko'paytmasi quyidagicha aniqlanadi:
Skalyar ko'paytmaning ta'rifi
Ba'zan bu amal nuqtali ko'paytma (inglizcha dot product) yoki chiziqli algebrada ichki ko'paytma (inner product) deb ham yuritiladi. Skalyar ko'paytmaning natijasi vektor emas, balki haqiqiy son (skalyar) ekanligiga e'tibor bering!
- \(\vec v = 5\vec i - 8\vec j,\,\,\vec w = \vec i + 2\vec j\)
- \(\vec a = \left\langle {0,3, - 7} \right\rangle ,\,\,\vec b = \left\langle {2,3,1} \right\rangle \)
Yechim
Formulaga mos komponentlarni ko'paytirib qo'shamiz:
a \(\vec v \cdot \vec w = (5)(1) + (-8)(2) = 5 - 16 = - 11\)
b \(\vec a \cdot \vec b = (0)(2) + (3)(3) + (-7)(1) = 0 + 9 - 7 = 2\)
Quyida skalyar ko'paytmaning asosiy xossalari keltirilgan.
Skalyar ko'paytmaning xossalari
Ushbu xossalarning isbotlari to'g'ridan-to'g'ri hisoblashga asoslanadi. Shuning uchun ulardan ikkitasini isbotlab ko'rsatamiz, qolganlarini mustaqil isbotlash uchun qoldiramiz.
\(\vec u \cdot \left( {\vec v + \vec w} \right) = \vec u \cdot \vec v + \vec u \cdot \vec w\) xossaning isboti
Uchta umumiy \(n\) o'lchamli vektor bilan boshlaymiz: \(\vec u = \left\langle {{u_1},{u_2}, \ldots ,{u_n}} \right\rangle \), \(\vec v = \left\langle {{v_1},{v_2}, \ldots ,{v_n}} \right\rangle \) va \(\vec w = \left\langle {{w_1},{w_2}, \ldots ,{w_n}} \right\rangle \). Teorema umumiy o'lcham uchun o'rinli bo'lgani sababli, isbotni ham \(n\) o'lchamli hol uchun keltiramiz:
Tenglikning chap tomonini hisoblaymiz:
\[\begin{align*}\vec u \cdot \left( {\vec v + \vec w} \right) & = \left\langle {{u_1},{u_2}, \ldots ,{u_n}} \right\rangle \cdot \left( {\left\langle {{v_1},{v_2}, \ldots ,{v_n}} \right\rangle + \left\langle {{w_1},{w_2}, \ldots ,{w_n}} \right\rangle } \right)\\ & = \left\langle {{u_1},{u_2}, \ldots ,{u_n}} \right\rangle \cdot \left\langle {{v_1} + {w_1},{v_2} + {w_2}, \ldots ,{v_n} + {w_n}} \right\rangle \\ & = {{u_1}\left( {{v_1} + {w_1}} \right)+{u_2}\left( {{v_2} + {w_2}} \right)+ \ldots +{u_n}\left( {{v_n} + {w_n}} \right)} \\ & = {{u_1}{v_1} + {u_1}{w_1} + {u_2}{v_2} + {u_2}{w_2} + \ldots + {u_n}{v_n} + {u_n}{w_n}} \\ & = \left( {{u_1}{v_1} + {u_2}{v_2} + \ldots + {u_n}{v_n}} \right) + \left( {{u_1}{w_1} + {u_2}{w_2} + \ldots + {u_n}{w_n}} \right) \\ & = \left\langle {{u_1},{u_2}, \ldots ,{u_n}} \right\rangle \cdot \left\langle {{v_1},{v_2}, \ldots ,{v_n}} \right\rangle + \left\langle {{u_1},{u_2}, \ldots ,{u_n}} \right\rangle \cdot \left\langle {{w_1},{w_2}, \ldots ,{w_n}} \right\rangle \\ & = \vec u \cdot \vec v + \vec u \cdot \vec w\end{align*}\]Isbot: Agar \(\vec v \cdot \vec v = 0\) bo'lsa, u holda \(\vec v = \vec 0\)
Bu juda sodda isbotdir. \(\vec v = \left\langle {{v_1},{v_2}, \ldots ,{v_n}} \right\rangle \) vektorining o'z-o'ziga skalyar ko'paytmasini yozaylik:
\[\begin{align*}\vec v \cdot \vec v & = \left\langle {{v_1},{v_2}, \ldots ,{v_n}} \right\rangle \cdot \left\langle {{v_1},{v_2}, \ldots ,{v_n}} \right\rangle \\ & = v_1^2 + v_2^2 + \cdots + v_n^2\\ & = 0\end{align*}\]Har qanday haqiqiy son uchun \(v_i^2 \ge 0\) ekanligi ma'lum bo'lgani sababli, manfiy bo'lmagan sonlar yig'indisi faqat va faqat har bir qo'shiluvchi alohida-alohida nol bo'lgandagina nolga teng bo'ladi, ya'ni \(v_i^2 = 0 \Rightarrow v_i = 0\). Bu esa barcha komponentlar nol ekanligini, binobarin \(\vec v = \vec 0\) ekanligini isbotlaydi.
Skalyar ko'paytma ajoyib geometrik ma'noga ham ega. Dastlab \(\vec a\) va \(\vec b\) vektorlari orasidagi burchak \(\theta\) bo'lsin, bunda \(0 \le \theta \le \pi \) (ya'ni \(0^\circ \le \theta \le 180^\circ\)), quyidagi chizmada ko'rsatilganidek:
U holda quyidagi asosiy geometrik teorema o'rinlidir.
Skalyar ko'paytmaning geometrik formulasi
Geometrik formulaning isboti
Yuqoridagi chizmani uchburchak hosil qiladigan holatga keltiraylik:
Ushbu uchta vektor \(AOB\) uchburchagini hosil qiladi. Uchburchak tomonlarining uzunliklari mos vektorlarning modullariga teng.
Trigonometriyadagi Kosinuslar teoremasiga ko'ra:
\[{\left\| {\vec a - \vec b} \right\|^2} = {\left\| {\vec a} \right\|^2} + {\left\| {\vec b} \right\|^2} - 2\left\| {\vec a} \right\|\,\,\left\| {\vec b} \right\|\cos \theta \]Ikkinchi tomondan, skalyar ko'paytmaning xossalaridan foydalanib, chap tomonni o'z-o'ziga ko'paytma sifatida yozishimiz mumkin:
\[{\left\| {\vec a - \vec b} \right\|^2} = (\vec a - \vec b) \cdot (\vec a - \vec b) = \vec a \cdot \vec a - 2(\vec a \cdot \vec b) + \vec b \cdot \vec b = {\left\| {\vec a} \right\|^2} - 2(\vec a \cdot \vec b) + {\left\| {\vec b} \right\|^2}\]Har ikkala tenglikni tenglashtirsak:
\[{\left\| {\vec a} \right\|^2} - 2(\vec a \cdot \vec b) + {\left\| {\vec b} \right\|^2} = {\left\| {\vec a} \right\|^2} + {\left\| {\vec b} \right\|^2} - 2\left\| {\vec a} \right\|\,\,\left\| {\vec b} \right\|\cos \theta \]\({\left\| {\vec a} \right\|^2}\) va \({\left\| {\vec b} \right\|^2}\) hadlar qisqarib ketadi, har ikkala tomonni \(-2\) ga bo'lsak:
\[\vec a \cdot \vec b = \left\| {\vec a} \right\|\,\,\left\| {\vec b} \right\|\cos \theta \]Teorema to'liq isbotlandi.
Ushbu teoremadagi formula amaliyotda ko'pincha skalyar ko'paytmani topish uchun emas, balki ikki vektor orasidagi burchakni topish uchun ishlatiladi:
\[\cos \theta = \frac{\vec a \cdot \vec b}{\left\| {\vec a} \right\|\,\,\left\| {\vec b} \right\|}\]Shuni unutmangki, chizma ikki o'lchamda tasvirlangan bo'lsa-da, teorema har qanday o'lchamli fazodagi vektorlar uchun qat'iy o'rinlidir.
Quyidagi misolni ko'raylik.
Yechim
Buning uchun skalyar ko'paytmani hamda har ikkala vektorning modullarini hisoblaymiz:
\[\vec a \cdot \vec b = (3)(0) + (-4)(5) + (-1)(2) = - 22\] \[\left\| {\vec a} \right\| = \sqrt {9 + 16 + 1} = \sqrt {26} \hspace{0.25in}\hspace{0.25in}\left\| {\vec b} \right\| = \sqrt {0 + 25 + 4} = \sqrt {29} \]U holda burchak kosinusi va burchakning o'zi:
\[\begin{align*}\cos \theta & = \frac{{\vec a \cdot \vec b}}{{\left\| {\vec a} \right\|\,\,\left\| {\vec b} \right\|}} = \frac{{ - 22}}{{\sqrt {26} \sqrt {29} }} \approx - 0.8011927\\ & \\ \theta & = \arccos\left( { - 0.8011927} \right) \approx 2.5{\mbox{ radian }} \approx 143.24^\circ\end{align*}\]Skalyar ko'paytma ikki vektorning perpendikulyar (ortogonal) ekanligini tekshirishning eng qulay va kuchli vositasidir. Matematikada perpendikulyarlik o'rniga ko'pincha ortogonal atamasi qo'llaniladi.
Ortogonal (Perpendikulyar) vektorlar
Agar ikki vektor ortogonal bo'lsa, ular orasidagi burchak \(\theta = 90^\circ\) (\(\frac{\pi}{2}\)) bo'ladi. \(\cos(90^\circ) = 0\) bo'lgani sababli, ikki nol bo'lmagan vektor ortogonal bo'lishi uchun ularning skalyar ko'paytmasi nolga teng bo'lishi zarur va yetarlidir:
\[\vec a \cdot \vec b = 0\]Xuddi shuningdek, agar ikki vektor parallel bo'lsa, ular orasidagi burchak yo \(0^\circ\) (yo'nalishdosh) yoki \(180^\circ\) (qarama-qarshi yo'nalgan) bo'ladi. Demak:
\[\vec a \cdot \vec b = \left\| {\vec a} \right\|\,\,\left\| {\vec b} \right\|\,\,\left( {\theta = 0^\circ } \right)\hspace{0.25in}{\mbox{YOKI}}\hspace{0.25in}\vec a \cdot \vec b = - \left\| {\vec a} \right\|\,\,\left\| {\vec b} \right\|\,\,\left( {\theta = 180^\circ } \right)\]- \(\vec a = \left\langle {6, - 2, - 1} \right\rangle ,\,\,\vec b = \left\langle {2,5,2} \right\rangle \)
- \(\displaystyle \vec u = 2\vec i - \vec j,\,\,\vec v = - \frac{1}{2}\vec i + \frac{1}{4}\vec j\)
a \(\vec a = \left\langle {6, - 2, - 1} \right\rangle ,\,\,\vec b = \left\langle {2,5,2} \right\rangle \) Yechim
Avval skalyar ko'paytmani tekshiramiz:
\[\vec a \cdot \vec b = (6)(2) + (-2)(5) + (-1)(2) = 12 - 10 - 2 = 0\]Skalyar ko'paytma nolga teng chiqdi. Demak, bu ikki vektor ortogonaldir.
b \(\displaystyle \vec u = 2\vec i - \vec j,\,\,\vec v = - \frac{1}{2}\vec i + \frac{1}{4}\vec j\) Yechim
Yana avval skalyar ko'paytmani hisoblaymiz:
\[\vec u \cdot \vec v = (2)\left(-\frac{1}{2}\right) + (-1)\left(\frac{1}{4}\right) = - 1 - \frac{1}{4} = - \frac{5}{4}\]Ko'paytma nol emas, demak ortogonal emas. Endi ularning modullarini tekshiramiz:
\[\left\| {\vec u} \right\| = \sqrt{2^2 + (-1)^2} = \sqrt 5 \hspace{0.25in}\left\| {\vec v} \right\| = \sqrt {\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2} = \sqrt {\frac{1}{4} + \frac{1}{16}} = \sqrt {\frac{5}{{16}}} = \frac{{\sqrt 5 }}{4}\]E'tibor bering:
\[\vec u \cdot \vec v = - \frac{5}{4} = - \sqrt 5 \left( {\frac{{\sqrt 5 }}{4}} \right) = - \left\| {\vec u} \right\|\,\,\left\| {\vec v} \right\|\]Shunday qilib, \(\vec u \cdot \vec v = - \|\vec u\| \|\vec v\|\), bu esa burchak \(180^\circ\) ekanligini, ya'ni vektorlar o'zaro parallel ekanligini ko'rsatadi (shuningdek, \(\vec v = -\frac{1}{4}\vec u\) munosabatdan ham buni darhol ko'rish mumkin edi).
Endi skalyar ko'paytmaning yana bir nechta ajoyib amaliy tatbiqlarini ko'rib chiqamiz.
Proyeksiyalar
Proyeksiyalarni tushunishning eng yaxshi yo'li chizmalarga qarashdir. Bizga \(\vec a\) va \(\vec b\) vektorlari berilgan bo'lsin. Biz \(\vec b\) vektorining \(\vec a\) vektoridagi proyeksiyasini topmoqchimiz. Bu proyeksiya \({{\mathop{\rm proj}\nolimits} _{\vec a}}\vec b\) belgisi bilan ifodalanadi. Quyidagi ikki holat bunga misoldir:
\(\vec b\) ning \(\vec a\) dagi proyeksiyasini olish uchun \(\vec b\) ning oxirgi nuqtasidan \(\vec a\) ga (yoki uning davomiga) to'g'ri burchak ostida perpendikulyar tushiramiz. Proyeksiya — bu ikkala vektor boshlangan umumiy nuqtadan boshlanib, tushirilgan perpendikulyar asosi bo'lgan nuqtada tugaydigan yangi vektordir. Ushbu vektor albatta \(\vec a\) ga parallel bo'ladi.
\(\vec b\) ning \(\vec a\) dagi proyeksiyasini hisoblash uchun ajoyib qulay formula mavjud:
\(\boldsymbol{\vec b}\) vektorining \(\boldsymbol{\vec a}\) dagi vektor proyeksiyasi
Shuni ta'kidlash joizki, proyeksiyaning yo'nalishiga qarab qaysi vektor qaysisiga proyeksiyalanayotganini aniq belgilash shart. Masalan, \(\vec a\) ning \(\vec b\) dagi proyeksiyasi butunlay boshqacha bo'ladi:
\(\boldsymbol{\vec a}\) vektorining \(\boldsymbol{\vec b}\) dagi vektor proyeksiyasi
Ko'rib turganingizdek, bu butunlay boshqa vektor bo'lib, u \(\vec b\) ga paralleldir. Shuning uchun indeksdagi vektorga doimo diqqat bilan qarang.
Buni misollarda ko'raylik.
Yechim
Bizga skalyar ko'paytma va \(\vec a\) modulining kvadrati zarur bo'ladi:
\[\vec a \cdot \vec b = (1)(2) + (0)(1) + (-2)(-1) = 2 + 0 + 2 = 4\hspace{0.25in}\hspace{0.25in}{\left\| {\vec a} \right\|^2} = 1^2 + 0^2 + (-2)^2 = 5\]U holda kerakli proyeksiya vektori:
\[\begin{align*}{{\mathop{\rm proj}\nolimits} _{\vec a}}\vec b & = \frac{{\vec a \cdot \vec b}}{{{{\left\| {\vec a} \right\|}^2}}}\vec a\\ & = \frac{4}{5}\left\langle {1,0, - 2} \right\rangle \\ & = \left\langle {\frac{4}{5},0, - \frac{8}{5}} \right\rangle \end{align*}\]Taqqoslash uchun, endi teskari holatni hisoblaymiz.
Yechim
Endi \(\vec b\) modulining kvadratini hisoblaymiz:
\[\vec a \cdot \vec b = 4\hspace{0.25in}\hspace{0.25in}{\left\| {\vec b} \right\|^2} = 2^2 + 1^2 + (-1)^2 = 6\]Proyeksiya esa:
\[\begin{align*}{{\mathop{\rm proj}\nolimits} _{\vec b}}\vec a & = \frac{{\vec a \cdot \vec b}}{{{{\left\| {\vec b} \right\|}^2}}}\vec b\\ & = \frac{4}{6}\left\langle {2,1, - 1} \right\rangle \\ & = \left\langle {\frac{4}{3},\frac{2}{3}, - \frac{2}{3}} \right\rangle \end{align*}\]Ko'rib turganimizdek, ikki proyeksiya mutlaqo boshqa-boshqa vektorlardir.
Yo'naltiruvchi kosinuslar
Skalyar ko'paytmaning ushbu tatbiqi aynan uch o'lchamli fazoga xosdir.
Uch o'lchamli fazodagi \(\vec a\) vektorini olaylik. Ushbu vektor musbat \(x\)-o'qi bilan \(\alpha\), musbat \(y\)-o'qi bilan \(\beta\) va musbat \(z\)-o'qi bilan \(\gamma\) burchaklarini hosil qiladi. Bu burchaklar yo'naltiruvchi burchaklar, ularning kosinuslari esa yo'naltiruvchi kosinuslar deb ataladi.
Quyidagi chizmada vektor va yo'naltiruvchi burchaklar tasvirlangan:
Yo'naltiruvchi kosinuslar quyidagi qulay formulalar orqali hisoblanadi:
Yo'naltiruvchi kosinuslar formulalari
bu yerda \(\vec i\), \(\vec j\) va \(\vec k\) standart bazis vektorlaridir.
Buni birinchi skalyar ko'paytma misolida tekshirib ko'raylik:
\[\vec a \cdot \,\vec i = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \cdot \left\langle {1,0,0} \right\rangle = {a_1}\]Yo'naltiruvchi kosinuslar haqida bir nechta ajoyib teoremalar mavjud:
Yo'naltiruvchi kosinuslarning xossalari
- \(\vec u = \left\langle {\cos \alpha ,\cos \beta ,\cos \gamma } \right\rangle \) vektori birlik vektor hisoblanadi.
- \({\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1\)
- \(\vec a = \left\| {\vec a} \right\|\left\langle {\cos \alpha ,\cos \beta ,\cos \gamma } \right\rangle \)
Yo'naltiruvchi kosinuslarga doir amaliy misol ko'raylik.
Yechim
Avval vektorning modulini topamiz:
\[\left\| {\vec a} \right\| = \sqrt {2^2 + 1^2 + (-4)^2} = \sqrt {4 + 1 + 16} = \sqrt {21} \]U holda yo'naltiruvchi kosinuslar va burchaklar quyidagicha hisoblanadi:
\[\begin{align*}\cos \alpha & = \frac{2}{{\sqrt {21} }}\hspace{0.25in}\hspace{0.25in}\alpha = \arccos\left(\frac{2}{\sqrt{21}}\right) \approx 1.119{\mbox{ radian}} \approx 64.12^\circ\\ \cos \beta & = \frac{1}{{\sqrt {21} }}\hspace{0.25in}\hspace{0.25in}\beta = \arccos\left(\frac{1}{\sqrt{21}}\right) \approx 1.351{\mbox{ radian}} \approx 77.40^\circ\\ \cos \gamma & = \frac{{ - 4}}{{\sqrt {21} }}\hspace{0.25in}\hspace{0.25in}\gamma = \arccos\left(\frac{-4}{\sqrt{21}}\right) \approx 2.632{\mbox{ radian}} \approx 150.79^\circ\end{align*}\]