Su'aalo Tusaale ah oo Ka Hadlaya Vektors Laba-cabbir ah Nidaamka Isku-dubaridka
Vektor waa tiro leh baaxad iyo jiho labadaba. Vektor waxaa si joogto ah loogu isticmaalaa mowduucyo kala duwan oo xisaabta iyo fiisigiska ah si ay u matalaan ifafaale kala duwan. Maqaalkan, waxaan ka wada hadli doonnaa tusaalooyin vektor laba-geesood ah oo ku jira nidaamka isku-dubaridka.
Fikradaha Aasaasiga ah ee Vektors-ka ee Nidaamyada Isku-dubaridka
Vektor ku jira nidaamka isku-dubbaridka laba-geesoodka ah waxaa loo matali karaa sida \(\vec{A} = (a_1, a_2)\), halkaasoo \(a_1\) uu yahay qaybta x ee vektorka iyo \(a_2\) uu yahay qaybta y ee vektorka. Vektorka waxaa loo kala saari karaa laba qaybood, kuwaas oo kala ah qaybta x iyo qaybta y.
Isugeynta iyo Kala-goynta Vektor-ka
Ku darista laba vektor \(\vec{A} = (a_1, a_2)\) iyo \(\vec{B} = (b_1, b_2)\) waa:
\[
\vec{A} + \vec{B} = (a_1 + b_1, a_2 + b_2)
\]
In kasta oo hoos u dhacu yahay:
\[
\vec{A} – \vec{B} = (a_1 – b_1, a_2 – b_2)
\]
Isku-dhufashada Cabbirka
Haddii \(\vec{A} = (a_1, a_2)\) iyo \(k\) ay yihiin scalar, markaas \(k\vec{A}\) waa:
\[
k\vec{A} = (k \cdot a_1, k \cdot a_2)
\]
Cabbirka Vektorka
Cabbirka ama dhererka vektorka \(\vec{A} = (a_1, a_2)\) waa:
\[
|\vec{A}| = \sqrt{a_1^2 + a_2^2}
\]
Vektor-ka Cutubka
Vektor cutub waa vektor dhererkiisu yahay hal cutub. Vektor cutubka \(\vec{A} = (a_1, a_2)\) waa:
\[
\hat{A} = \frac{\vec{A}}{|\vec{A}|} = \left( \frac{a_1}{\sqrt{a_1^2 + a_2^2}}, \frac{a_2}{\sqrt{a_1^2 + a_2^2}} \right)
\]
Su'aalo iyo Doodo Tusaale ah
Su'aal 1aad: Isugeynta iyo Kala-goynta Vektorrada
Laba vector ayaa loo bixiyay sidan soo socota: \(\vec{A} = (3, 4)\) iyo \(\vec{B} = (1, 2)\). Soo hel badeecada \(\vec{A} + \vec{B}\) iyo \(\vec{A} – \vec{B}\).
Dood:
\[
\vec{A} + \vec{B} = (3 + 1, 4 + 2) = (4, 6)
\]
\[
\vec{A} – \vec{B} = (3 – 1, 4 – 2) = (2, 2)
\]
Su'aal 2: Isku-dhufashada Miisaanka
Marka la eego vektorka \(\vec{C} = (2, -3)\), xisaabi \(3\vec{C}\) iyo \(-2\vec{C}\).
Dood:
\[
3\vec{C} = 3 \cdot (2, -3) = (6, -9)
\]
\[
-2\vec{C} = -2 \cdot (2, -3) = (-4, 6)
\]
Su'aal 3: Cabbirka Vektor-ka
Xisaabi baaxadda vektorka \(\vec{D} = (5, 12)\).
Dood:
\[
|\vec{D}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
\]
Su'aal 4: Vektorrada Cutubyada
Soo hel vektor-ka cutubka ee vektor-ka \(\vec{E} = (4, 3)\).
Dood:
\[
|\vec{E}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]
\[
\hat{E} = \frac{\vec{E}}{|\vec{E}|} = \left( \frac{4}{5}, \frac{3}{5} \right)
\]
Su'aal 5: Goobta iyo Masaafada Vektorrada
Laba dhibcood oo ku jira diyaaradda isku-dhafka laba-geesoodka ah waa P(2, 3) iyo Q(5, 7). Go'aami vector-ka booska laga bilaabo barta P ilaa barta Q iyo masaafada u dhaxaysa.
Dood:
Vektorka booska laga bilaabo P ilaa Q waa:
\[
\vec{PQ} = \vec{Q} – \vec{P} = (5 – 2, 7 – 3) = (3, 4)
\]
Masaafada u dhaxaysa dhibcaha P iyo Q waa:
\[
|\vec{PQ}| = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
Su'aal 6: Natiijada Badeecada Dhibcaha
Haddii \(\vec{F} = (-3, 4)\) iyo \(\vec{G} = (2, 1)\), xisaabi natiijada dhibcaha ee \(\vec{F} \cdot \vec{G}\).
Dood:
Badeecada dhibcaha ee laba vectors waa:
\[
\vec{F} \cdot \vec{G} = (-3) \cdot 2 + 4 \cdot 1 = -6 + 4 = -2
\]
Su'aal 7: Xagal u dhexeeya laba Vektor
Haddii \(\vec{H} = (7, -4)\) iyo \(\vec{I} = (3, 0)\), go'aami xagasha u dhaxaysa labada vector.
Dood:
Si loo go'aamiyo xagasha u dhaxaysa laba vector, waxaan isticmaalnaa qaacidada:
\[
\cos \theta = \frac{\vec{H} \cdot \vec{I}}{|\vec{H}| |\vec{I}|}
\]
Marka hore, xisaabi badeecada dhibcaha \(\vec{H} \cdot \vec{I}\):
\[
\vec{H} \cdot \vec{I} = 7 \cdot 3 + (-4) \cdot 0 = 21 + 0 = 21
\]
Kadib, xisaabi baaxadaha \(\vec{H}\) iyo \(\vec{I}\):
\[
|\vec{H}| = \sqrt{7^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}
\]
\[
|\vec{I}| = \sqrt{3^2 + 0^2} = \sqrt{9} = 3
\]
Ku qor qiimayaashan qaacidada:
\[
\cos \theta = \frac{21}{\sqrt{65} \cdot 3} = \frac{21}{3\sqrt{65}} = \frac{7}{\sqrt{65}}
\]
Markaa, \(\theta = \cos^{-1} \left( \frac{7}{\sqrt{65}} \right) \).
Su'aal 8: Saadaalinta Vektor-ka
Wixii vektorrada \(\vec{J} = (2, 1)\) iyo \(\vec{K} = (-1, 3)\), xisaabi saadaasha \(\vec{J}\) ee \(\vec{K}\).
Dood:
Saadaasha \(\vec{J}\) ee ku saabsan \(\vec{K}\) waa:
\[
\text{proj}_{\vec{K}} \vec{J} = \bidix( \frac{\vec{J} \cdot \vec{K}}{|\vec{K}|^2} \right) \vec{K}
\]
Marka hore, xisaabi badeecada dhibcaha \(\vec{J} \cdot \vec{K}\):
\[
\vec{J} \cdot \vec{K} = 2 \cdot (-1) + 1 \cdot 3 = -2 + 3 = 1
\]
Kadib, baaxadda \(\vec{K}\):
\[
|\vec{K}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}
\]
Sidaas darteed,:
\[
|\vec{K}|^2 = 10
\]
Geli qaacidada:
\[
\text{proj}_{\vec{K}} \vec{J} = \left( \frac{1}{10} \right) \vec{K} = \left( \frac{1}{10} \right) (-1, 3) = \left( -\frac{1}{10}, \frac{3}{10} \right)
\]
Kuwani waa tusaalooyin dhibaatooyin iyo doodo la xiriira vectors laba-geesood ah oo ku jira nidaamka isku-dubaridka. Faham wanaagsan oo ku saabsan vectors-ka ayaa waxtar u yeelan kara codsiyo badan oo ku saabsan xisaabta, fiisigiska, iyo injineernimada. Ku celcelinta tusaalooyin kala duwan waxay sii xoojin kartaa fahamkaaga fikraddan, taasoo u oggolaanaysa in si wax ku ool ah loogu dabaqo xaalado kala duwan.
