Co-loidhneachd

Nuair a bhios tu ag obrachadh ann an trì meudan, chan eil ach aon dòigh air a dhearbhadh gu bheil trì puingean air an aon loidhne (co-loidhneach) agus 's e sin sealltainn gu bheil cùrsa coitcheann ann. Feumaidh tu bheactoran a chleachdadh airson sin.

Seo mar a shealladh tu gu bheil \(A(4,1,3)\), \(B(8,4,6)\) agus \(C(20,13,15)\) co-loidhneach.

An toiseach, tagh dà phìos loidhne le cùrsa comharraichte agus puing choitcheann:

\[\overrightarrow {AB} = \left( \begin{array}{l} 4\\ 3\\ 3 \end{array} \right),\,\overrightarrow {BC} = \left( \begin{array}{l} 12\\ \,\,9\\ \,\,9 \end{array} \right)\]

Sgrìobh aon mar iomad dhen fhear eile:

\(\overrightarrow {BC} = 3\left( \begin{array}{l} 4\\ 3\\ 3 \end{array} \right)\), me \(\overrightarrow {BC} = 3 \times \overrightarrow {AB}\)

agus sgrìobh co-dhùnadh.

Mar sin tha cùrsa coitcheann aig \(\overrightarrow {AB}\) agus \(\overrightarrow {BC}\).

Cuir crìoch air an dearbhadh.

Tha puing choitcheann aig \(\overrightarrow {AB}\) agus \(\overrightarrow {BC}\) Mar sin tha \(A\), \(B\) agus \(C\) co-loidhneach.

Question

Ma tha \(\overrightarrow {PR} = \left( \begin{array}{l} \,\,\,\,\,5\\ - 1\\- 2 \end{array} \right),\,\overrightarrow {QR} = \left( \begin{array}{l} - 5\\ \,\,\,\,\,1\\ \,\,\,\,\,2 \end{array} \right)\) seall gu bheil \(P\),\(Q\) agus \(R\) co-loidhneach.

\(\overrightarrow{QR}=- 1\left(\begin{array}{l}\,\,\,\,\,5\\- 1\\- 2\end{array}\right)\), sin \(\overrightarrow {QR} = - 1 \times \overrightarrow {PR}\)

Mar sin tha an aon chùrsa aig \(\overrightarrow {QR}\) agus \(\overrightarrow {PR}\).

Tha puing choitcheann aig \(\overrightarrow {QR}\) agus \(\overrightarrow {PR}\) sin \(R\).

Mar sin tha \(P\), \(Q\) agus \(R\) co-loidhneach.