Respuesta :
Answer:
[tex] a.b = (1*0) + (-4*2) +(1*-2)= -10[/tex]
And replacing we got:
[tex] cos \theta = \frac{-10}{\sqrt{18} \sqrt{8}}= -\frac{10}{\sqrt{144}}= -\frac{10}{12}= -\frac{5}{6}[/tex]
And we can find the angle with the inverse cosine function and we got:
[tex]\theta =cos^{-1} (-\frac{5}{6})= 146.44°[/tex]
Step-by-step explanation:
for this case we can use the following identity:
[tex] cos \theta = \frac{a.b}{|a| |b|}[/tex]
We can begin finding the norm for each vector and we got:
[tex] |a| =\sqrt{(1)^2 +(-4)^2 +(1)^2}= \sqrt{18}[/tex]
[tex] |b| =\sqrt{(0)^2 +(2)^2 +(-2)^2}= \sqrt{8}[/tex]
Now we can find the dot product and we got:
[tex] a.b = (1*0) + (-4*2) +(1*-2)= -10[/tex]
And replacing we got:
[tex] cos \theta = \frac{-10}{\sqrt{18} \sqrt{8}}= -\frac{10}{\sqrt{144}}= -\frac{10}{12}= -\frac{5}{6}[/tex]
And we can find the angle with the inverse cosine function and we got:
[tex]\theta =cos^{-1} (-\frac{5}{6})= 146.44°[/tex]
