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

`\theta = cos^(-1) ((10)/(√(9) √(25)))=cos^(-1) ((10)/(15)) = cos^(-1) ((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`

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|= √((1)^2 +(2)^2 +(-2)^2)=√(9) =3`

`|b| =√((4)^2 +(0)^2 +(-3)^2)=√(25)= 5`

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

`cos \theta = (ab)/(|a| |b|)`

And the angle is given by:

`\theta = cos^(-1) ((ab)/(|a| |b|))`

If we replace we got:

`\theta = cos^(-1) ((10)/(√(9) √(25)))=cos^(-1) ((10)/(15)) = cos^(-1) ((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`