Co-phàirtean bheactor
A' cur-ris bheactoran
Tha na co-phàirtean \((\frac{2}{5})\) aig \(\overrightarrow {PQ}\)
Tha na co-phàirtean \((\frac{4}{-3})\) aig \(\overrightarrow {QR}\)
Le bhith a' cur-ris nam bheactoran seo, gheibh sinn \(\overrightarrow {PQ} + \overrightarrow {QR} = \overrightarrow {PR}\)
'S e an riaghailt airson a bhith a' cur-ris nan co-phàirtean:
\(\left( \begin{array}{l} a\\ b \end{array} \right) + \left( \begin{array}{l} c\\ d \end{array} \right) = \left( \begin{array}{l} a + c\\ b + d \end{array} \right)\)
Mar sin tha \(\overrightarrow {PQ} + \overrightarrow {QR} = \overrightarrow {PR}\) a' coimhead mar seo:
\(\left( \begin{array}{l}2\\5\end{array} \right) + \left( \begin{array}{l}4\\-3\end{array} \right) = \left( \begin{array}{l}6\\2\end{array} \right)\)
A' toirt-air-falbh bheactoran
Nuair a bhios tu a' toirt-air-falbh bheactor, tha e an aon rud ri bhith a' cur-ris dreach àicheil dhen bheactor (cuimhnich nuair a nì thu bheactor àicheil gum bi an cùrsa aige a' dol an taobh eile).
\(\left( \begin{array}{l} a\\ b \end{array} \right) - \left( \begin{array}{l} c\\ d \end{array} \right) = \left( \begin{array}{l} a - c\\ b - d \end{array} \right)\)
Coimhead air an diagram agus smaoinich air a bhith a' dol bho X gu Z. Ciamar a sgrìobhadh tu an t-slighe ann a' bheactoran a' cleachdadh nam beactoran \(\overrightarrow {XY}\) agus \(\overrightarrow {ZY}\) a-mhàin?
Dh'fhaodadh tu a ràdh gur e bheactor \(\overrightarrow {XY}\) a th' ann le gluasad air ais air \(\overrightarrow {ZY}\).
Mar sin faodaidh sinn an t-slighe bho X gu Z a sgrìobhadh mar:
\(\overrightarrow {XY} - \overrightarrow {ZY} = \overrightarrow {XZ}\)
Tha e a' coimhead mar seo ann an àireamhan:
\(\left( \begin{array}{l} 4\\ 2 \end{array} \right) - \left( \begin{array}{l} 1\\ 2 \end{array} \right) = \left( \begin{array}{l} 3\\ 0 \end{array} \right)\)
Feuch a-nis na ceistean gu h-ìosal.
Question
Ma tha \(x = \left( \begin{array}{l} 1\\ 3 \end{array} \right),y = \left( \begin{array}{l} - 2\\ 4 \end{array} \right)agus\,z = \left( \begin{array}{l} - 1\\ - 2 \end{array} \right)\), obraich a-mach:
- \(- y\)
- \(x - y\)
- \(2x + 3z\)
- \(\left( \begin{array}{l} 2\\ - 4 \end{array} \right)\) (An do dh'atharraich thu na soidhnichean?)
- \(\left( \begin{array}{l} 1\\ 3 \end{array} \right) - \left( \begin{array}{l} - 2\\ 4 \end{array} \right) = \left( \begin{array}{l} 1 - - 2\\ 3 - 4 \end{array} \right) = \left( \begin{array}{l} 3\\ - 1 \end{array} \right)\)
- \(2\left( \begin{array}{l} 1\\ 3 \end{array} \right) + 3\left( \begin{array}{l} - 1\\ - 2 \end{array} \right) = \left( \begin{array}{l} 2\\ 6 \end{array} \right) + \left( \begin{array}{l} - 3\\ - 6 \end{array} \right) = \left( \begin{array}{l} - 1\\ 0 \end{array} \right)\)
Question
Airson bheactoran \(u = \left( {\begin{array}{*{20}{c}} 2\\ 5\\ 9 \end{array}} \right)agus\,v = \left( {\begin{array}{*{20}{c}} 7\\ 3\\ { - 4} \end{array}} \right)\)
obraich a-mach, u + v.
u + v \(= \left( {\begin{array}{*{20}{c}} 2\\ 5\\ 9 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 7\\ 3\\ { - 4} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {2 + 7}\\ {5 + 3}\\ {9 + ( - 4)} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 9\\ 8\\ 5 \end{array}} \right)\)