Bheactoran suidheachaidh
Airson puing P, canaidh sinn gur e a' bheactor bhon origin gu puing P bheactor suidheachadh P.
Nuair a tha na co-chomharran (1,4,8) aig P, tha na co-phàirtean \(\left( \begin{array}{l}1\\4\\8\end{array} \right)\) aig bheactor suidheachadh P.
Canaidh sinn gu bheil \(\overrightarrow {OP}=\left( \begin{array}{l}1\\4\\8\end{array} \right)\)
Faodaidh sinn bheactoran suidheachaidh a chleachdadh gus co-phàirtean bheactor obrachadh a-mach.
Eisimpleir
Tha na co-chomharran (1,4,8) aig P agus tha na co-chomharran (-3,1,-4) aig Q.
Obraich a-mach co-phàirtean bheactor \(\overrightarrow {PQ}\).
Freagairt
Tha \(\overrightarrow {OP}=\left( \begin{array}{l}1\\4\\8\end{array} \right)\) agus \(\overrightarrow {OQ}=\left( \begin{array}{l}-3\\1\\-4\end{array} \right)\) againn.
Nuair a bhios sinn a' smaoineachadh air bheactor obrachadh a-mach, faodaidh sinn sgeidse a dhèanamh. 'S dòcha nach bi càil a choltas aig an sgeidse ri diagram mhionaideach, agus faodaidh e fiù 's a bhith 2sh an àite 3sh.
Faodaidh sinn a ràdh gu bheil \(\overrightarrow {PQ} = \overrightarrow {PO}+ \overrightarrow {OQ}\)
Cuimhnich gur e \(\overrightarrow {PO}\) an t-àicheil aig \(\overrightarrow {OP}\)
Mar sin \(\overrightarrow {PQ}=\left( \begin{array}{l}-1\\-4\\-8\end{array} \right)+\left( \begin{array}{l}-3\\1\\-4\end{array} \right)\)
\(\overrightarrow {PQ}= \left( \begin{array}{l}-4\\-3\\-12\end{array} \right)\)
Meud
Tha am meud ag innse dè cho mòr 's a tha a' bheactor.
Ma tha \(\overrightarrow {PQ} = \left( {\begin{array}{*{20}{c}} a\\ b\\ c \end{array}} \right)\) tha meud \(\overrightarrow {PQ}\) air a sgrìobhadh mar \(\left| {\overrightarrow {PQ} } \right|\) agus tha e air obrachadh a-mach leis an fhoirmle:
\(\left| {\overrightarrow {PQ} } \right| = \sqrt {{a^2} + {b^2} + {c^2}}\)
Feuch a-nis na ceistean gu h-ìosal.
Question
Obraich a-mach meud a' bheactor \(p=\left( \begin{array}{l}3\\7\end{array} \right)\)
\(\left | p \right |=\sqrt{3^{2}+7^{2}}\)
\(\left | p \right |=\sqrt{9+49}\)
\(\left | p \right |=\sqrt{58}\)
Question
Ma tha \(\overrightarrow {PQ} = \left( {\begin{array}{*{20}{c}} 5\\ 4\\ { - 2} \end{array}} \right)\) obraich a-mach a mheud.
\(\left| {\overrightarrow {PQ} } \right| = \sqrt {{5^2} + {4^2} + {{( - 2)}^2}} = \sqrt {45} = 3\sqrt 5 \,aonadan\)
Question
Ma tha \(a = \left( {\begin{array}{*{20}{c}} 5\\ { - 1}\\ 2 \end{array}} \right)\,agus\,b = \left( {\begin{array}{*{20}{c}} 7\\ 9\\ { - 2} \end{array}} \right)\) obraich a-mach \(\left| {\overrightarrow {AB} } \right|\)
\(\left| {\overrightarrow {AB} } \right| = \left( {\begin{array}{*{20}{c}} 7\\ 9\\ { - 2} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 5\\ { - 1}\\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {7 - 5}\\ {9 - ( - 1)}\\ { - 2 - 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2\\ {10}\\ { - 4} \end{array}} \right)\)
Mar sin:
\(\left| {\overrightarrow {AB} } \right| = \sqrt {{2^2} + {{10}^2} + {{( - 4)}^2}} = \sqrt {120} = 2\sqrt {30} \,aonadan\)
