Bheactoran co-ionann

Ma tha an aon mheudachd agus an aon chùrsa aig dà bheactor tha sin a' ciallachadh gu bheil iad co-ionann, ge bith dè an suidheachadh aca.

Two vectors with the same magnitude and direction are equal

A' cur-ris bheactoran

Nuair a bhios sinn a' cur-ris bheactoran, bidh sinn a' leantainn na riaghailt:

\[\left( \begin{array}{l} a\\ b \end{array} \right) + \left( \begin{array}{l} c\\ d \end{array} \right) = \left( \begin{array}{l} a + c\\ b + d \end{array} \right)\]

Seall ris a' ghraf gu h-ìosal agus chì thu na gluasadan eadar PQ, QR agus PR.

Vectors PQ and QR share point Q

Tha bheactor \(\overrightarrow {PQ}\) air a leantainn le bheactor \(\overrightarrow {QR}\) a' sealltainn a' ghluasaid bho P gu R.

\[\overrightarrow {PQ} + \overrightarrow {QR} = \overrightarrow {PR}\]

Seo mar a tha a' bheactor nuair a tha e air a sgrìobhadh:

\[\left( \begin{array}{l}2\\5\end{array} \right) + \left( \begin{array}{l}\,\,\,\,\,4\\- 3\end{array} \right) = \left( \begin{array}{l}6\\2\end{array} \right)\]

A' toirt-air-falbh bheactoran

Tha toirt-air-falbh an aon rud ri bhith a' cur-ris riochd àicheil dhen bheactor (cuimhnich nuair a nì thu bheactor àicheil gum bi an cùrsa aige a' dol an taobh eile).

\[\left( \begin{array}{l} a\\ b \end{array} \right) - \left( \begin{array}{l} c\\ d \end{array} \right) = \left( \begin{array}{l} a - c\\ b - d \end{array} \right)\]

Vectors XY and ZY both end at point Y

Coimhead air an diagram agus smaoinich gu bheil thu a' dol bho X gu Z. Ciamar a sgrìobhadh tu an t-slighe ann a' bheactoran a' cleachdadh nam bheactoran \(\overrightarrow {XY}\) 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:

\[\overrightarrow {XY} - \overrightarrow {ZY} = \overrightarrow {XZ}\]

Seo e sgrìobhte ann an àireamhan:

\[\left( \begin{array}{l} 4\\ 2 \end{array} \right) - \left( \begin{array}{l} 1\\ 2 \end{array} \right) = \left( \begin{array}{l} 3\\ 0 \end{array} \right)\]

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. \[2x + 3z\]

  1. \(\left( \begin{array}{l}\,\,\,\,\,2\\- 4\end{array} \right)\) An do dh'atharraich thu na samhlan?
  2. \[\left( \begin{array}{l}\,1\\3\end{array} \right) - \left( \begin{array}{l}- 2\\\,\,\,\,4\end{array} \right) = \left( \begin{array}{l}\,1 - - 2\\3 - \,\,\,\,4\end{array} \right) = \left( \begin{array}{l}\,\,\,\,\,3\\- 1\end{array} \right)\]
  3. \[\left( \begin{array}{l}\,1\\3\end{array} \right) + 3\left( \begin{array}{l}- 1\\- 2\end{array} \right) = \left( \begin{array}{l}2\\6\end{array} \right) + \left( \begin{array}{l}- 3\\- 6\end{array} \right) = \left( \begin{array}{l}- 1\\\,\,\,\,0\end{array}\right)\]