Su'aal Tusaale ah oo ka hadlaysa Ku Darista Laba Vektor iyadoo la adeegsanayo Habka Parallelogram-ka
Ku darista Vektorka waa fikrad muhiim ah oo ku jirta fiisigiska iyo xisaabta, oo inta badan loo isticmaalo in lagu qeexo dhacdooyinka dabiiciga ah iyo dhibaatooyinka nolol maalmeedka. Waxaa jira dhowr hab oo lagu daro laba vektor, mid ka mid ah waa habka parallelogram. Habkani ma aha oo kaliya mid dareen leh laakiin sidoo kale wuxuu bixiyaa muuqaal xooggan oo ku saabsan sida laba vektor isugu daraan si ay u sameeyaan vektor natiijo leh. Maqaalkan, waxaan ku eegi doonaa dhowr tusaale oo ku saabsan ku darista vektorka iyadoo la adeegsanayo habka parallelogram, iyo xalalkooda.
Waa maxay Vektor?
Kahor inta aynaan guda gelin dhibaatooyinka tusaalaha ah, waxaan u baahanahay inaan fahanno qeexitaanka aasaasiga ah ee vektor. Vektor waa tiro leh labadaba baaxad (dherer) iyo jihayn. Tusaalooyinka caadiga ah ee vektorrada waxaa ka mid ah xawaaraha, dardargelinta, xoogga, iyo barokaca. Vektor waxaa loo matali karaa qaybihiisa (i, j, k) isku-duwayaasha Cartesian ama dhererkiisa iyo jihadiisa (xagalkiisa).
Habka Isbarbardhigga
Habka parallelogram-ku waa hal hab oo lagu daro laba vector. Habkan, waxaan u matalaynaa laba vectors laba dhinac oo parallelogram ah. Vector-ka natiijada leh waa geesoodka parallelogram-ka oo ka bilaabmaya barta bilowga labada vector. Xisaab ahaan, haddii aan haysanno laba vectors \(\vec{A}\) iyo \(\vec{B}\), natiijada ka dhalata waa \( \vec{R} = \vec{A} + \vec{B} \).
Habka tallaabo-tallaabo ah ee loo adeegsado habka parallelogram waa sida soo socota:
1. Ka soo sawir vektorka \(\vec{A}\) barta laga bilaabo.
2. Laga bilaabo dhammaadka vector-ka \(\vec{A}\), sawir vector-ka \(\vec{B}\).
3. Sawir xariiq barbar socota vektorka \(\vec{B}\) laga bilaabo barta bilowga \(\vec{A}\).
4. Sawir xariiq barbar socota vector \(\vec{A}\) laga bilaabo dhammaadka vector \(\vec{B}\).
5. Sawir googoos ah laga bilaabo barta bilowga ilaa geeska ka soo horjeeda si aad u hesho vektor-ka natiijada leh \(\vec{R}\).
Su'aalo iyo Doodo Tusaale ah
Su'aal 1aad
Ka soo qaad inaan haysanno laba vectors \(\vec{A}\) iyo \(\vec{B}\):
– \(\vec{A}\) wuxuu leeyahay dherer (magnitude) oo ah 5 cutub iyo jiho ah 0° (ama dhinaca dhidibka x ee togan),
– \(\vec{B}\) wuxuu leeyahay dherer 3 cutub ah iyo jiho 90° ah (ama dhidibka y-ga togan).
Waa maxay qiimaha ka dhasha ku darista labadan vectors iyadoo la adeegsanayo habka parallelogram-ka?
Dood:
1. Sawir vektor \(\vec{A}\) oo ku socda dhidibka x ee togan oo leh dherer dhan 5 cutub.
2. Laga bilaabo dhammaadka vektorka \(\vec{A}\), sawir vektorka \(\vec{B}\) oo ku teedsan dhidibka y-ga togan oo leh dherer dhan 3 cutub.
3. Laga bilaabo barta bilowga \(\vec{A}\), sawir xariiq barbar socota \(\vec{B}\).
4. Laga bilaabo dhammaadka \(\vec{B}\), sawir xariiq barbar socota \(\vec{A}\).
5. Natiijadu waa barbarlelogram leh qaab-dhismeed toosan oo ah vector-ka natiijada \(\vec{R}\).
Maadaama \(\vec{A}\) iyo \(\vec{B}\) ay isku toosan yihiin, waxaan isticmaali karnaa aragtida Pythagorean si aan u xisaabinno dhererka vector-ka natiijada ka dhashay:
\[ R = \sqrt{A^2 + B^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \qiyaastii 5.83 \]
Jihada vektor-ka natiijada leh waxaa lagu xisaabin karaa iyadoo la adeegsanayo trigonometry. Haddii \(\theta\) uu yahay xagasha u dhaxaysa natiijada iyo \(\vec{A}\):
\[ \tan(\theta) = \frac{B}{A} = \frac{3}{5} \]
sidaas darteed:
\[ \theta = \tan^{-1}\left(\frac{3}{5}\right) \qiyaastii 30.96^\circle \]
Sidaas darteed, vektor-ka natiijada leh \(\vec{R}\) wuxuu leeyahay cabbir dhan 5.83 cutub iyo jiho dhan 30.96° laga bilaabo \(\vec{A}\).
Su'aal 2aad
Laba vektor \(\vec{C}\) iyo \(\vec{D}\) ayaa loo bixiyay sidan soo socota:
– \(\vec{C}\) oo leh dherer 4 cutub ah iyo jiho 45° ah.
– \(\vec{D}\) oo leh dherer 6 cutub ah iyo jiho 120° ah.
Go'aami vektor-ka natiijada ka soo baxda \(\vec{R}\) marka la isku daro labada vektor.
Dood:
Si aad ugu darto laba vector oo aan isku toosanayn ama qaabab kala duwan leh, waxaad isticmaali kartaa qaybaha Cartesian.
1. U kala jejebi \(\vec{C}\) iyo \(\vec{D}\) qaybaha x iyo y.
Wixii \(\vec{C}\):
\[ C_x = C \cos(45^\circ) = 4 \cos(45^\circ) = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2} \qiyaastii 2.83 \]
\[ C_y = C \sin(45^\circ) = 4 \sin(45^\circ) = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2} \qiyaastii 2.83 \]
Wixii \(\vec{D}\):
\[ D_x = D \cos(120^\circle) = 6 \cos(120^\circle) = 6 \cdot (-\frac{1}{2}) = -3 \]
\[ D_y = D \sin(120^\circ) = 6 \sin(120^\circ) = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} \qiyaastii 5.20 \]
2. Ku dar qaybaha x iyo y ee labada vector:
\[ R_x = C_x + D_x = 2.83 + (-3) = -0.17 \]
\[ R_y = C_y + D_y = 2.83 + 5.20 = 8.03 \]
3. Xisaabi baaxadda iyo jihada vektor-ka natiijada leh \(\vec{R}\):
\[ R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-0.17)^2 + 8.03^2} = \sqrt{0.03 + 64.48} = \sqrt{64.51} \qiyaastii 8.03 \]
\[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{8.03}{-0.17}\right) \approx \tan^{-1}(-47.24) \]
Maadaama natiijadu ay tahay taban, waxaan ku darnaa 180° si aan u helno xagasha nidaamka rubuca saxda ah:
\[ \theta \qiyaastii \tan^{-1}(47.24) + 180^\circle \qiyaastii 271.93^\circle \]
Markaa, vektor-ka natiijada leh \(\vec{R}\) wuxuu leeyahay cabbir dhan 8.03 cutub iyo jiho dhan 271.93°, ama waxaan dhihi karnaa qiyaastii 91.93° laga bilaabo dhidibka x ee taban ee rubuca afraad.
Xiritaanka
Habka parallelogram waa hab wax ku ool ah oo muuqaal ah oo lagu daro laba vectors. In kasta oo habkani u ekaan karo mid fudud vectors-ka fudud, waxaa muhiim ah in la fahmo in vectors-ka aadka u adag, badanaa waxaan u baahanahay inaan isticmaalno qaybaha Cartesian iyo farsamooyinka aljebrada ee horumarsan si aan u helno natiijooyin sax ah. Waxaan rajeyneynaa, tusaalooyinka kor ku xusan waxay bixiyaan sawir cad oo ku saabsan sida habkan loogu dabaqi karo xaalado kala duwan.
