Respuesta :
Answer:
[tex] \theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190[/tex]
Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel
And the anfle is approximately [tex] \theta \approx 48[/tex]
Step-by-step explanation:
For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.
a=[1,2,-2], b=[4,0,-3,]
The dot product on this case is:
[tex] a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10[/tex]
Since the dot product is not equal to zero then the two vectors are not orthogonal.
Now we can calculate the magnitude of each vector like this:
[tex] |a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3[/tex]
[tex] |b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5[/tex]
And finally we can calculate the angle between the vectors like this:
[tex] cos \theta = \frac{ab}{|a| |b|}[/tex]
And the angle is given by:
[tex] \theta = cos^{-1} (\frac{ab}{|a| |b|})[/tex]
If we replace we got:
[tex] \theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190[/tex]
Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel
And the anfle is approximately [tex] \theta \approx 48[/tex]
