An toradh scalair
Tha an toradh scalair \(a.b\) air a mhìneachadh mar \(\textbf{a.b}=\left|\textbf{a}\right|\left|\textbf{b}\right|\cos\theta \) far an e \(\theta\) an ceàrn eadar \(\textbf{a}\) agus \(\textbf{b}\).
Bhon mhìneachadh seo, chì sinn cuideachd gu bheil \(\textbf{a.b} = {a_x}{b_x} + {a_y}{b_y} + {a_z}{b_z}\)
'S e am prìomh fheum aig an toradh scalair an ceàrn \(\theta\) obrachadh a-mach.
\(\left|\textbf{a}\right|\left|\textbf{b}\right|\cos \theta = \textbf{a.b}\)
Mar sin tha \(\cos \theta = \frac{{\textbf{a.b}}}{{\left|\textbf{a}\right|\left|\textbf{b}\right|}}\) far a bheil \(\textbf{a.b} = {a_x}{b_x} + {a_y}{b_y} + {a_z}{b_z}\)
Eisimpleir
Obraich a-mach an ceàrn \(\theta\) air an diagram gu h-ìosal.
Sgrìobh an riaghailt a tha thu a' cleachdadh airson na ceist seo:
\(\cos \theta = \frac{{p.q}}{{\left| p \right|\left| q \right|}}\)
Obraich a-mach \(p.q\)
\({p_x}{q_x} + {p_y}{q_y} + {p_z}{q_z} =\)
\(3 \times 2 + ( - 1) \times 4 + 4 \times 2\)
\(= 10\)
Obraich a-mach \(\left| p \right|\) agus \(\left| q \right|\)
\(\left| p \right| = \sqrt {9 + 1 + 16} = \sqrt {26}\)
\(\left| q \right| = \sqrt {4 + 16 + 4} = \sqrt {24}\)
Ionadaich:
\(\cos \theta = \frac{{10}}{{\sqrt {26} \sqrt {24} }} = 0.400\)
Obraich a-mach \(\theta\)
\(\theta = 66.4^\circ\)
Ma tha do fhreagairt àicheil aig a' cheum ionadachaidh, tha an ceàrn eadar \(90^\circ\) agus \(180^\circ\). Mar eisimpleir, ma tha \(\cos \theta = - 0.362\) tha \(\theta = 111^\circ\)
