Vectors ba lambobi ne na yau da kullun ba, don haka ba za a iya amfani da su kai tsaye ba. Dole ne mu yi amfani da ninka vector. Akwai nau'ikan ninka vector guda biyu: ninka digo da ninka giciye. Ana kuma kiran ninka digo scalar saboda yana samar da adadi na scalar. Ana kuma kiran ninka giciye ventral multiplication saboda yana samar da adadi na vector. Misali, akwai vector guda biyu, wato vector. A dan B. Ƙirƙirar ƙwayoyin cuta masu sikelin A dan B an bayyana tare da AB KTunda fagen yana amfani da alamar ɗigo, ana kiran wannan ninkawa samfurin digo. Yawan vector na A dan B an bayyana tare da A x BDomin yana amfani da rubutu x, to wannan ninkawa ana kiransa ninkawa giciye.
Misali, idan aka yi la'akari da vector A dan B kamar yadda aka nuna a hoton da ke ƙasa. Samfurin ɗigo tsakanin vectors A dan B an rubuta kamar AB (A wuri B).
Don ayyana samfurin digo na vectors A dan B (AB), vector da aka nuna A da kuma vectors Ta hanyar ywanda ke samar da kusurwa θ. Na gaba za mu zana hasashen vector B zuwa ga alkiblar vector AWannan hasashen wani ɓangare ne na vector B wanda yake daidai da vector A, wanda girmansa iri ɗaya ne da B cos θ.
Don haka, mun ayyana AB a matsayin babban vector A ninka ta hanyar abubuwan vector B wanda yake daidai da AA fannin lissafi za mu iya rubuta shi kamar haka:
AB cos θ lamba ce ta yau da kullun (scalar). Saboda haka, ana kiran samfurin digo samfurin scalar. Me zai faru idan samfurin digo tsakanin vectors A dan B aka juya zuwa BA kafin mu fayyace BADa farko mun zana hasashen vector A zuwa vectors B (duba hoton da ke ƙasa).
Dangane da wannan hoton, za mu iya bayyana BA a matsayin babban vector B ninka ta hanyar abubuwan vector A wanda yake daidai da BA fannin lissafi za mu iya rubuta shi kamar haka:
Sakamakon samfurin ɗigo AB = AB cos θ da kuma sakamakon samfurin ɗigon BA = BA cos θDomin kuwa AB cos θ = BA cos θ, sannan ya shafi AB = BA
Wasu abubuwa game da ninka maki da kuke buƙatar sani:
1. Samfurin ɗigo ya cika dokar canja wuri.
AB = BA
2. Samfurin ɗigo ya cika dokar rarrabawa.
A. (B + C) = AB + AC
3. Idan vectors A da B sun daidaita da juna, to samfurin digo AB = 0
Lokacin da vector A dan B Daidai da juna, sannan kusurwar da aka kafa ita ce 90o. Cos 90o = 0. Saboda haka: AB = AB cos inna = AB kowa 90o = 0. A gefe guda kuma, BA = BA cos inna = BA kowa 90o = 0
4. Idan vector A da vector B suna cikin alkibla ɗaya , to AB = AB cos 0 o = AB
Lokacin da vector A dan B a cikin wannan alkibla, to kusurwar da aka kafa ita ce 0o. cos 0 = 1. Don haka, AB = AB cos inna = AB kowa 0o = ABAkasin haka BA = BA cos inna = BA kowa 0o = BA
(Bai kamata a ruɗe ka da AB dan BABabba AB = babba BAMisali, girman vector A = 2. girman vector B = 3. sannan AB = 2.3 = 6; wannan iri ɗaya ne da BA = 3.2 = 6.
5. Wani yanayi na vector guda biyu a cikin alkibla ɗaya, idan A = B to za mu sami AA = A 2 ko BB = B 2
6. Idan vectors A da B suna cikin alkibla daban-daban (lokacin da vectors guda biyu suna cikin alkibla daban-daban, kusurwar da aka samar ita ce 180º) , to sakamakon ninka AB = AB cos 180º = AB (-1) = -AB.
Kos 180º = -1.
Misalin matsalar:
Vektor A yana da girman raka'a 4 kuma vektor B yana da girman raka'a 3. Kayyade samfurin digo na vektor guda biyu idan kusurwoyin da vektor guda biyu suka samar sune 60º, 90º da 180º.
Tattaunawa
Tunda AB = BA , za mu iya zaɓar amfani da ɗayansu. Misali, muna amfani da AB
AB = AB cos theta
Girman A = raka'a 4 da girman B = raka'a 3.
