Question

find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B=[4,0,-3]

Answer 1

Answer:

$\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190$

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 $\theta \approx 48$

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:

$a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10$

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:

$|a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3$

$|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{ab}{|a| |b|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{ab}{|a| |b|})$

If we replace we got:

$\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190$

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 $\theta \approx 48$