Tambayoyi Misali Game da Ƙarin Vector ta Bangare
Ƙarin Vector wata hanya ce ta asali a fannin kimiyyar lissafi da lissafi da ake amfani da ita don nemo sakamakon vector guda biyu ko fiye. Hanyar warware ƙarin vector ta hanyar amfani da sassan jiki hanya ce mai matuƙar amfani, musamman lokacin da ake mu'amala da vectors a girma biyu ko uku. Wannan labarin zai yi bayani game da manufar ƙarin vector ta hanyar sassan jiki da kuma samar da misalai da dama na matsaloli da mafita.
Manufar Ƙarin Vector Mai Haɗawa
Kowace vector a cikin sararin girma biyu (2D) za a iya raba ta zuwa sassa biyu: bangaren x (kwance) da bangaren y (tsaye). A cikin girma uku (3D), vectors suna da ƙarin sashi, bangaren z (zurfi).
A ce muna da vectors guda biyu A da B. Ana iya bayyana sassan waɗannan vectors kamar haka:
– Vector A yana da sassa \(A_x\) da \(A_y\) a cikin 2D (ko kuma \(A_z\) a cikin 3D).
– Vector B yana da sassa \(B_x\) da \(B_y\) a cikin 2D (ko kuma \(B_z\) a cikin 3D).
Ƙara waɗannan vectors guda biyu zai samar da vector R mai sakamako wanda ke da waɗannan abubuwan:
\[ R_x = A_x + B_x \]
\[ R_y = A_y + B_y \]
Ga vectors a cikin 3D, bangaren z shima yana kamar haka:
\[ R_z = A_z + B_z \]
Bayan ƙididdige kowanne ɓangare na vector da aka samu, za mu iya samun modulus (girma) da alkiblar vector da aka samu ta amfani da dabarar:
\[ |R| = \sqrt{R_x^2 + R_y^2} \] (na 2D)
Ko kuma don 3D:
\[ |R| = \sqrt{R_x^2 + R_y^2 + R_z^2} \]
Kuma ana iya tantance alkiblar vector ɗin da ya haifar ta hanyar kusurwar da ke kan gatari masu daidaitawa.
Tambayoyi da Tattaunawa Samfura
Tambaya ta 1
An ba da vectors guda biyu a cikin jirgin sama mai girma biyu:
– A shine \(5 \, \text{unit}\) zuwa gabas.
– B shine \(3 \, \text{unit}\) a arewa.
Ƙayyade vector ɗin da aka samu sakamakon R.
Tattaunawa
Da farko, muna canza vector zuwa sassansa.
– Vector A: \(A = (5, 0)\) saboda yana da ɓangaren x kawai.
– Vector B: \(B = (0, 3)\) saboda yana da bangaren y kawai.
Ga jimlar abubuwan da aka haɗa:
\[ R_x = A_x + B_x = 5 + 0 = 5 \]
\[ R_y = A_y + B_y = 0 + 3 = 3 \]
Sannan sakamakon vector R shine:
\[ R = (5, 3) \]
Don ƙididdige tsawon (modulus) na vector R:
\[ |R| = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \kimanin 5.83 \]
Ana iya ƙididdige alkiblar vector R ta amfani da kusurwar θ zuwa ga axis ɗin x:
\[ \tan(\theta) = \frac{R_y}{R_x} = \frac{3}{5} \]
\[ \theta = \arctan\left(\frac{3}{5}\right) \kimanin 30.96^\circ \]
Saboda haka, vector mai sakamakon R yana da tsawon kimanin raka'a 5.83 kuma yana samar da kusurwar 30.96° tare da axis ɗin x.
Tambaya ta 2
An ba da vectors guda biyu a cikin girma uku:
– A shine \(3\hat{i} + 2\hat{j} + 1\hat{k}\)
– B shine \(1\hat{i} + 4\hat{j} + 2\hat{k}\)
Ƙayyade vector ɗin da aka samu sakamakon R.
Tattaunawa
Da farko, mun gano abubuwan da ke cikin kowanne vector:
– Vector A: \(A_x = 3\), \(A_y = 2\), \(A_z = 1\).
– Vector B: \(B_x = 1\), \(B_y = 4\), \(B_z = 2\).
Ga jimlar abubuwan da aka haɗa:
\[ R_x = A_x + B_x = 3 + 1 = 4 \]
\[ R_y = A_y + B_y = 2 + 4 = 6 \]
\[ R_z = A_z + B_z = 1 + 2 = 3 \]
Sannan sakamakon vector R shine:
\[ R = (4, 6, 3) \]
Don ƙididdige tsawon (modulus) na vector R:
\[ |R| = \sqrt{4^2 + 6^2 + 3^2} = \sqrt{16 + 36 + 9} = \sqrt{61} \kimanin 7.81 \]
Ana iya ƙididdige alkiblar vector R dangane da gatari x, y, da z ta amfani da cosine na darektan:
\[ \cos(\alpha) = \frac{R_x}{|R|} = \frac{4}{7.81} \kimanin 0.512 \]
\[ \alpha = \arccos(0.512) \kimanin 59.50^\cir \]
\[ \cos(\beta) = \frac{R_y}{|R|} = \frac{6}{7.81} \kimanin 0.768 \]
\[ \beta = \arccos(0.768) \kimanin 39.50^\cir \]
\[ \cos(\gamma) = \frac{R_z}{|R|} = \frac{3}{7.81} \kimanin 0.384 \]
\[ \gamma = \arccos(0.384) \kimanin 67.64^\circ \]
Saboda haka, vector mai sakamakon R yana da tsawon kimanin raka'a 7.81 kuma alkiblarsa dangane da axes x, y, da z sune 59.50°, 39.50°, da 67.64°.
Tambaya ta 3
An ba da vectors guda biyu:
– P yana da girman raka'a 4 kuma yana samar da kusurwar 45° zuwa ga axis ɗin x mai kyau.
– Q yana da girman raka'a 6 kuma yana samar da kusurwar 120° zuwa ga axis ɗin x mai kyau.
Ƙayyade vector ɗin da aka samu sakamakon R.
Tattaunawa
Da farko, mun raba vector zuwa sassan x da y:
– Vector P: \(P_x = 4\cos(45^\circ) = 4 \cdot \frac{\sqrt{2}}{2} \approx 2.83\), \(P_y = 4\sin(45^\circ) = 4 \cdot \frac{\sqrt{2}}{2} \approx 2.83\).
– Vector Q: \(Q_x = 6\cos(120^\circ) = 6 \cdot \left(-\frac{1}{2}\right) = -3\), \(Q_y = 6\sin(120^\circ) = 6 \cdot \frac{\sqrt{3}}{2} \approx 5.2\).
Ga jimlar abubuwan da aka haɗa:
\[ R_x = P_x + Q_x = 2.83 – 3 = -0.17 \]
\[ R_y = P_y + Q_y = 2.83 + 5.2 = 8.03 \]
Sannan, sakamakon vector R shine:
\[ R = (-0.17, 8.03) \]
Don ƙididdige tsawon (modulus) na vector R:
\[ |R| = \sqrt{(-0.17)^2 + 8.03^2} = \sqrt{0.0289 + 64.48} = \sqrt{64.509} \kimanin 8.03 \]
Alkiblar vector R:
\[ \tan(\theta) = \frac{R_y}{R_x} = \frac{8.03}{-0.17} = -47.24 \]
\[ \theta = \arctan(-47.24) \kimanin -88.99^\circle \]
Duk da haka, ana auna wannan kusurwar ne a kusa da axis ɗin x mara kyau, don haka ainihin kusurwar da ke cikin mahallin matsalar ita ce:
\[ 180^\circa – 88.99^\circa \kimanin 91.01^\circa \]
Saboda haka, vector mai sakamakon R yana da tsawon kimanin raka'a 8.03 kuma yana samar da kusurwar 91.01° tare da axis mai kyau na x.
Wannan labarin ya tattauna ƙarin vector ta hanyar amfani da sassan, yana ba da misalai da mafita da dama. Hanyar amfani da sassan tana da matuƙar amfani wajen sauƙaƙa lissafi da kuma samar da hanyar da ta dace don magance matsalolin vector a cikin ma'aunin lissafi na sararin samaniya.
