Su'aalo Tusaale ah oo Ka Hadlaya Ku Darista Vektor-ka Qayb ahaan
Ku darista Vektor waa hab aasaasi ah oo ku jira fiisigiska iyo xisaabta oo loo isticmaalo in lagu helo natiijada laba ama in ka badan vektor. Habka qaybaha ee lagu xallinayo ku darista vektor waa hab gaar ah oo faa'iido leh, gaar ahaan marka lala macaamilayo vektor-yada laba ama saddex cabbir. Maqaalkani wuxuu sharxi doonaa fikradda ku darista vektor-yada qaybaha wuxuuna bixin doonaa dhowr dhibaato iyo xalal tusaale ah.
Fikradda Ku Darista Vektor-ka Qaybaysan
Vektor kasta oo ku jira booska laba-cabbir ah (2D) waxaa loo qaybin karaa laba qaybood: qayb x (horizontal) ah iyo qayb y (toosan). Saddex cabbir (3D), vektorradu waxay leeyihiin qayb dheeraad ah, qaybta z (qoto dheer).
Ka soo qaad inaan haysanno laba vektor A iyo B. Qaybaha vektor-yadan waxaa loo qeexi karaa sidan soo socota:
– Vektor A wuxuu leeyahay qaybo \(A_x\) iyo \(A_y\) oo ku jira 2D (ama sidoo kale \(A_z\) oo ku jira 3D).
– Vektor B wuxuu leeyahay qaybo \(B_x\) iyo \(B_y\) oo ku jira 2D (ama sidoo kale \(B_z\) oo ku jira 3D).
Ku darista labadan vector waxay soo saari doontaa vector R oo natiijo leh oo leh qaybaha soo socda:
\[ R_x = A_x + B_x \]
\[ R_y = A_y + B_y \]
Vektorrada ku jira 3D, qaybta z sidoo kale waa kuwan soo socda:
\[ R_z = A_z + B_z \]
Ka dib marka aan xisaabino qayb kasta oo ka mid ah vektor-ka natiijada leh, waxaan ka heli karnaa modulus (magnitude) iyo jihada vektor-ka natiijada leh annagoo adeegsanayna qaacidada:
\[ |R| = \sqrt{R_x^2 + R_y^2} \] (loogu talagalay 2D)
Ama 3D:
\[ |R| = \sqrt{R_x^2 + R_y^2 + R_z^2} \]
Jihada vektor-ka natiijada leh waxaa lagu go'aamin karaa xagasha dhidibka isku-dhafka ah.
Su'aalo iyo Doodo Tusaale ah
Su'aal 1aad
Marka la eego laba vectors oo ku jira diyaarad laba-cabbir ah:
– A waa \(5 \, \text{unit}\) dhanka bari.
– B waa \(3 \, \text{unit}\) dhanka waqooyi.
Go'aami vektor-ka natiijada ka soo baxday R.
Dood
Marka hore, waxaan u beddelnaa vektor-ka qaybaha uu ka kooban yahay.
– Vektor A: \(A = (5, 0)\) sababtoo ah waxay leedahay qayb x ah oo keliya.
– Vektor B: \(B = (0, 3)\) sababtoo ah waxay leedahay qayb y ah oo keliya.
Waa kan wadarta qaybaha:
\[ R_x = A_x + B_x = 5 + 0 = 5 \]
\[ R_y = A_y + B_y = 0 + 3 = 3 \]
Markaas vektor-ka natiijada R waa:
\[ R = (5, 3) \]
Si loo xisaabiyo dhererka (modulus) ee vektorka R:
\[ |R| = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \qiyaastii 5.83 \]
Jihada vektorka R waxaa lagu xisaabin karaa iyadoo la isticmaalayo xagasha θ ilaa dhidibka x:
\[ \tan(\theta) = \frac{R_y}{R_x} = \frac{3}{5} \]
\[ \theta = \arctan\left(\frac{3}{5}\right) \qiyaastii 30.96^\circle \]
Sidaas darteed, vektor-ka natiijada ka soo baxda R wuxuu leeyahay dherer qiyaastii 5.83 unug wuxuuna sameeyaa xagal 30.96° ah oo leh dhidibka x.
Su'aal 2aad
Marka la eego laba vector oo saddex cabbir ah:
– A waa \(3\hat{i} + 2\hat{j} + 1\hat{k}\)
– B waa \(1\hat{i} + 4\hat{j} + 2\hat{k}\)
Go'aami vektor-ka natiijada ka soo baxday R.
Dood
Marka hore, waxaan aqoonsannaa qaybaha vektor kasta:
– Vektor A: \(A_x = 3\), \(A_y = 2\), \(A_z = 1\).
– Vektor B: \(B_x = 1\), \(B_y = 4\), \(B_z = 2\).
Waa kan wadarta qaybaha:
\[ R_x = A_x + B_x = 3 + 1 = 4 \]
\[ R_y = A_y + B_y = 2 + 4 = 6 \]
\[ R_z = A_z + B_z = 1 + 2 = 3 \]
Markaas vektor-ka natiijada R waa:
\[ R = (4, 6, 3) \]
Si loo xisaabiyo dhererka (modulus) ee vektorka R:
\[ |R| = \sqrt{4^2 + 6^2 + 3^2} = \sqrt{16 + 36 + 9} = \sqrt{61} \qiyaastii 7.81 \]
Jihada vektor-ka R marka loo eego dhidibyada x, y, iyo z waxaa lagu xisaabin karaa iyadoo la isticmaalayo cosine-ka agaasimaha:
\[ \cos(\alpha) = \frac{R_x}{|R|} = \frac{4}{7.81} \qiyaastii 0.512 \]
\[ \alpha = \arccos(0.512) \qiyaastii 59.50^\circle \]
\[ \cos(\beta) = \frac{R_y}{|R|} = \frac{6}{7.81} \qiyaastii 0.768 \]
\[ \beta = \arccos(0.768) \qiyaastii 39.50^\circle \]
\[ \cos(\gamma) = \frac{R_z}{|R|} = \frac{3}{7.81} \qiyaastii 0.384 \]
\[ \gamma = \arccos(0.384) \qiyaastii 67.64^\circle \]
Sidaas darteed, vektor-ka natiijada ka soo baxda R wuxuu leeyahay dherer dhan 7.81 cutub, jihooyinkiisuna waxay u dhigmaan dhidibyada x, y, iyo z waa 59.50°, 39.50°, iyo 67.64°.
Su'aal 3aad
Marka la eego laba vector:
– P waxay leedahay baaxad dhan 4 cutub waxayna samaysaa xagal 45° ah oo u jirta dhidibka x-ga ee togan.
– Q waxay leedahay baaxad dhan 6 cutub waxayna samaysaa xagal 120° ah oo u jirta dhidibka x-ga ee togan.
Go'aami vektor-ka natiijada ka soo baxday R.
Dood
Marka hore, waxaan u kala qaybinaynaa vektor-ka qaybaha x iyo y:
– Vektor P: \(P_x = 4\cos(45^\circ) = 4 \cdot \frac{\sqrt{2}}{2} \approx 2.83\), \(P_y = 4\sin(45^\circ) = 4 \cdot \frac{\sqrt{2}}{2} \approx 2.83\).
– Vektor Q: \(Q_x = 6\cos(120^\circ) = 6 \cdot \left(-\frac{1}{2}\right) = -3\), \(Q_y = 6\sin(120^\circ) = 6 \cdot \frac{\sqrt{3}}{2} \qiyaastii 5.2\).
Waa kan wadarta qaybaha:
\[ R_x = P_x + Q_x = 2.83 – 3 = -0.17 \]
\[ R_y = P_y + Q_y = 2.83 + 5.2 = 8.03 \]
Kadib, vector-ka natiijada R waa:
\[ R = (-0.17, 8.03) \]
Si loo xisaabiyo dhererka (modulus) ee vektorka R:
\[ |R| = \sqrt{(-0.17)^2 + 8.03^2} = \sqrt{0.0289 + 64.48} = \sqrt{64.509} \qiyaastii 8.03 \]
Jihada vektor-ka R:
\[ \tan(\theta) = \frac{R_y}{R_x} = \frac{8.03}{-0.17} = -47.24 \]
\[ \theta = \arctan(-47.24) \qiyaastii -88.99^\circle \]
Si kastaba ha ahaatee, xagashani waxaa lagu cabiraa dhidibka x ee taban, markaa xagasha dhabta ah ee macnaha guud ee dhibaatadu waa:
\[ 180^\circle – 88.99^\circle \qiyaastii 91.01^\circle \]
Sidaas darteed, vektor-ka natiijada ka soo baxda R wuxuu leeyahay dherer qiyaastii 8.03 unug wuxuuna sameeyaa xagal 91.01° ah oo leh dhidibka x ee togan.
Maqaalkani wuxuu ka hadlay ku darista vector-ka qaybaha, isagoo bixinaya dhowr tusaale oo dhibaatooyin iyo xalal ah. Habka qaybaha ku salaysan ayaa aad waxtar ugu leh fududeynta xisaabinta iyo bixinta hab nidaamsan oo lagu xalliyo dhibaatooyinka vector-ka ee cabbirka xisaabta ee booska.
