Piv txwv ntawm cov lus nug sib tham txog kev ua haujlwm vector
Cov haujlwm vector yog lub tswv yim tseem ceeb hauv kev lej uas feem ntau tshwm sim hauv ntau qhov kev tshawb fawb, xws li physics, engineering, thiab computer science. Hauv tsab xov xwm no, peb yuav tham txog ntau qhov piv txwv ntawm cov haujlwm vector thiab lawv cov kev daws teeb meem los muab kev nkag siab tob dua thiab meej dua. Cov piv txwv no yuav npog cov haujlwm yooj yim xws li kev ntxiv thiab rho tawm vector, nrog rau cov haujlwm siab dua xws li kev sib npaug scalar thiab kev sib npaug hla-vector.
1. Kev Ntxiv thiab Rho Vector
Piv txwv lus nug 1
Muab ob lub vectors A thiab B hauv daim ntawv tivthaiv:
\[ \mathbf{A} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix} \]
\[ \mathbf{B} = \begin{pmatrix} -1 \\ 4 \\ 2 \end{pmatrix} \]
Xam qhov tshwm sim ntawm kev ntxiv thiab rho tawm ntawm ob lub vectors.
Kev Sib Tham
Rau kev ntxiv vector, peb ntxiv txhua qhov sib xws ntawm ob lub vectors.
\[ \mathbf{A} + \mathbf{B} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix} + \begin{pmatrix} -1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 + (-1) \\ 3 + 4 \\ -1 + 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 1 \end{pmatrix} \]
Rau kev rho tawm vector, peb rho tawm txhua qhov sib xws ntawm ob qho vectors.
\[ \mathbf{A} – \mathbf{B} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix} – \begin{pmatrix} -1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 – (-1) \\ 3 – 4 \\ -1 – 2 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \\ -3 \end{pmatrix} \]
2. Kev Sib Npaug Scalar los ntawm Vector
Piv txwv lus nug 2
Muab ib qho vector C thiab ib qho scalar k:
\[ \mathbf{C} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} \]
\[ k = 4 \]
Xam cov khoom scalar ntawm vector C los ntawm scalar k.
Kev Sib Tham
Kev sib npaug ntawm scalar los ntawm vector yog ua los ntawm kev sib npaug txhua feem ntawm vector los ntawm scalar.
\[ k \mathbf{C} = 4 \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \cdot 1 \\ 4 \cdot (-2) \\ 4 \cdot 3 \end{pmatrix} = \begin{pmatrix} 4 \\ -8 \\ 12 \end{pmatrix} \]
3. Khoom Muag
Piv txwv lus nug 3
Muab ob lub vectors D thiab E:
\[ \mathbf{D} = \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix} \]
\[ \mathbf{E} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \]
Xam cov dot product ntawm ob lub vectors.
Kev Sib Tham
Cov dot product ntawm ob lub vectors yog tau los ntawm kev ntxiv cov khoom ntawm lawv cov khoom sib xws.
\[ \mathbf{D} \cdot \mathbf{E} = 3 \cdot 1 + (-2) \cdot 0 + 4 \cdot (-1) = 3 + 0 - 4 = -1 \]
4. Khoom Sib Txawv
Piv txwv lus nug 4
Muab ob lub vectors F thiab G:
\[ \mathbf{F} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} \]
\[ \mathbf{G} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \]
Xam qhov sib txawv ntawm ob lub vectors.
Kev Sib Tham
Qhov khoom sib tshuam ntawm ob lub vectors hauv qhov chaw peb-seem yog tau los ntawm kev siv tus determinant ntawm lub matrix tsim los ntawm cov vectors ntawd. Qhov khoom sib tshuam yog muab los ntawm cov mis:
\[ \mathbf{F} \times \mathbf{G} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 1 & -1 & 2 \end{vmatrix} \]
Qhov no tuaj yeem suav tau raws li hauv qab no:
\[
\mathbf{F} \times \mathbf{G} = \mathbf{i} \begin{vmatrix} 3 & 4 \\ -1 & 2 \end{vmatrix} – \mathbf{j} \begin{vmatrix} 2 & 4 \\ 1 & 2 \end{vmatrix} + k \end{vmatrix} + k 3 \\ 1 & -1 \end{vmatrix}
\]
Xam tus determinant ntawm txhua submatrix:
\[
= \mathbf{i} (3 \cdot 2 – 4 \cdot -1) – \mathbf{j} (2 \cdot 2 – 4 \cdot 1) + \mathbf{k} (2 \cdot -1 – 3 \cdot 1)
\]
\[
= \mathbf{i} (6 + 4) – \mathbf{j} (4 – 4) + \mathbf{k} (-2 – 3)
\]
\[
= \mathbf{i} (10) – \mathbf{j} (0) + \mathbf{k} (-5)
\]
\[
= \begin{pmatrix} 10 \\ 0 \\ -5 \end{pmatrix}
\]
Yog li, qhov sib txawv ntawm F thiab G yog:
\[ \mathbf{F} \times \mathbf{G} = \begin{pmatrix} 10 \\ 0 \\ -5 \end{pmatrix} \]
5. Txheeb xyuas lub kaum sab xis ntawm ob lub vectors
Piv txwv lus nug 5
Muab ob lub vectors H thiab kuv:
\[ \mathbf{H} = \begin{pmatrix} 6 \\ 2 \\ 3 \end{pmatrix} \]
\[ \mathbf{I} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} \]
Txheeb xyuas lub kaum sab xis ntawm ob lub vectors.
Kev Sib Tham
Lub kaum sab xis \(\theta\) ntawm ob lub vectors tuaj yeem nrhiav tau los ntawm kev siv kev sib raug zoo ntawm cov khoom dot thiab qhov loj ntawm ob lub vectors:
\[ \mathbf{H} \cdot \mathbf{I} = \| \mathbf{H} \| \| \mathbf{I} \| \cos \theta \]
Ua ntej, xam cov khoom dot \( \mathbf{H} \cdot \mathbf{I} \):
\[ \mathbf{H} \cdot \mathbf{I} = 6 \cdot 1 + 2 \cdot 4 + 3 \cdot (-2) = 6 + 8 – 6 = 8 \]
Tom ntej no, xam qhov loj ntawm ob qho vectors:
\[ \| \mathbf{H} \| = \sqrt{6^2+2^2+3^2} = \sqrt{36+4+9} = \sqrt{49} = 7 \]
\[ \| \mathbf{I} \| = \sqrt{1^2 + 4^2 + (-2)^2} = \sqrt{1+16+4} = \sqrt{21} \]
Tom qab ntawd, hloov cov nqi no rau hauv cov qauv kaum sab xis:
\[ \cos \theta = \frac{\mathbf{H} \cdot \mathbf{I}}{\| \mathbf{H} \| \| \mathbf{I} \|} = \frac{8}{7\sqrt{21}} \]
\[ \theta = \cos^{-1} \left( \frac{8}{7\sqrt{21}} \right) \]
Yog li qhov kawg, peb tuaj yeem siv lub tshuab xam zauv los nrhiav tus nqi ntawm lub kaum sab xis:
\[ \theta \approx 73,4^\circ \]
Xaus
Lub tswv yim ntawm kev ua haujlwm vector yog qhov tseem ceeb hauv kev lej thiab kev tshawb fawb. Tsab xov xwm no tham txog ntau qhov teeb meem piv txwv thiab lawv cov kev daws teeb meem, xws li kev ntxiv thiab rho tawm vector, scalar multiplication, dot product, cross product, thiab kev txiav txim siab lub kaum sab xis ntawm ob lub vectors. Los ntawm kev ua haujlwm los ntawm cov piv txwv no, peb vam tias yuav txhim kho koj txoj kev nkag siab txog kev ua haujlwm vector thiab pab koj daws cov teeb meem uas cuam tshuam nrog vectors hauv ntau qhov xwm txheej.
