2.3 The dot product  (Page 3/16)

1. To find the cosine of the angle formed by the two vectors, substitute the components of the vectors into [link] :
   
   $\begin{array}{cc}\hfill \text{cos}\theta & =\frac{(\text{i}+\text{j}+\text{k})·\left(2\text{i}-\text{j}-3\text{k}\right)}{\Vert \text{i}+\text{j}+\text{k}\Vert ·\Vert 2\text{i}-\text{j}-3\text{k}\Vert }\hfill \\ & =\frac{1(2)+(1)(\mathrm{-1})+(1)(\mathrm{-3})}{\sqrt{1^2+1^2+1^2}\sqrt{2^2+{(\mathrm{-1})}^2+{(\mathrm{-3})}^2}}\hfill \\ & =\frac{\mathrm{-2}}{\sqrt{3}\sqrt{14}}=\frac{\mathrm{-2}}{\sqrt{42}}.\hfill \end{array}$
   
   Therefore, $\theta =\text{arccos}\frac{\mathrm{-2}}{\sqrt{42}}$ rad.
2. Start by finding the value of the cosine of the angle between the vectors:
   
   $\begin{array}{cc}\hfill \text{cos}\theta & =\frac{⟨2,5,6⟩·⟨\mathrm{-2},\mathrm{-4},4⟩}{\Vert ⟨2,5,6⟩\Vert ·\Vert ⟨\mathrm{-2},\mathrm{-4},4⟩\Vert }\hfill \\ & =\frac{2(\mathrm{-2})+(5)(\mathrm{-4})+(6)(4)}{\sqrt{2^2+5^2+6^2}\sqrt{{(\mathrm{-2})}^2+{(\mathrm{-4})}^2+4^2}}\hfill \\ & =\frac{0}{\sqrt{65}\sqrt{36}}=0.\hfill \end{array}$
   
   Now, $\text{cos}\theta =0$ and $0\le \theta \le \pi ,$ so $\theta =\pi \text{/}2.$

Using the definition, we need only check the dot product of the vectors:

Because $\text{p}·\text{q}=0,$ the vectors are orthogonal ( [link] ).

1. Let α be the angle formed by v and i:
   
   $\begin{array}{cc}\hfill \text{cos}\alpha & =\frac{\text{v}·\text{i}}{\Vert \text{v}\Vert ·\Vert \text{i}\Vert }\hfill \\ & =\frac{⟨2,3,3⟩·⟨1,0,0⟩}{\sqrt{2^2+3^2+3^2}\sqrt{1}}\hfill \\ & =\frac{2}{\sqrt{22}}.\hfill \end{array}$
   
   
   $\alpha =\text{arccos}\frac{2}{\sqrt{22}}\approx 1.130\text{rad}.$
2. Let β represent the angle formed by v and j :
   
   $\begin{array}{cc}\hfill \text{cos}\beta & =\frac{\text{v}·\text{j}}{\Vert \text{v}\Vert ·\Vert \text{j}\Vert }\hfill \\ & =\frac{⟨2,3,3⟩·⟨0,1,0⟩}{\sqrt{2^2+3^2+3^2}\sqrt{1}}\hfill \\ & =\frac{3}{\sqrt{22}}.\hfill \end{array}$
   
   
   $\beta =\text{arccos}\frac{3}{\sqrt{22}}\approx 0.877\text{rad.}$
3. Let γ represent the angle formed by v and k :
   
   $\begin{array}{cc}\hfill \text{cos}\gamma & =\frac{\text{v}·\text{k}}{\Vert \text{v}\Vert ·\Vert \text{k}\Vert }\hfill \\ & =\frac{⟨2,3,3⟩·⟨0,0,1⟩}{\sqrt{2^2+3^2+3^2}\sqrt{1}}\hfill \\ & =\frac{3}{\sqrt{22}}.\hfill \end{array}$
   
   
   $\gamma =\text{arccos}\frac{3}{\sqrt{22}}\approx 0.877\text{rad.}$