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)

Diagram of arrow vectors

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)

Diagram of arrow vectors

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:

  1. - y
  2. x - y
  3. 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)