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 $$\left| q \right|$$

$\left| p \right| = \sqrt {9 + 1 + 16} = \sqrt {26}$

$\left| q \right| = \sqrt {4 + 16 + 4} = \sqrt {24}$

$\cos \theta = \frac{{10}}{{\sqrt {26} \sqrt {24} }} = 0.400$
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 $$\theta = 111^\circ$$