Answer:
a) ![u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168](https://tex.z-dn.net/?f=%20u%20v%3D%20%28-3%29%2A%284%29%20%2B%20%289%29%2A%28-12%29%2B%20%286%29%2A%28-8%29%3D-168)
Since the dot product is not equal to zero then the two vectors are not orthogonal.
![|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}](https://tex.z-dn.net/?f=%20%7Cu%7C%3D%20%5Csqrt%7B%28-3%29%5E2%20%2B%289%29%5E2%20%2B%286%29%5E2%7D%3D%5Csqrt%7B126%7D)
![|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}](https://tex.z-dn.net/?f=%20%7Cv%7C%20%3D%5Csqrt%7B%284%29%5E2%20%2B%28-12%29%5E2%20%2B%28-8%29%5E2%7D%3D%5Csqrt%7B224%7D)
![cos \theta = \frac{uv}{|u| |v|}](https://tex.z-dn.net/?f=%20cos%20%5Ctheta%20%3D%20%5Cfrac%7Buv%7D%7B%7Cu%7C%20%7Cv%7C%7D)
![\theta = cos^{-1} (\frac{uv}{|u| |v|})](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20cos%5E%7B-1%7D%20%28%5Cfrac%7Buv%7D%7B%7Cu%7C%20%7Cv%7C%7D%29)
If we replace we got:
![\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20cos%5E%7B-1%7D%20%28%5Cfrac%7B-168%7D%7B%5Csqrt%7B126%7D%20%5Csqrt%7B224%7D%7D%29%3Dcos%5E%7B-1%7D%20%28-1%29%20%3D%20%5Cpi)
Since the angle between the two vectors is 180 degrees we can conclude that are parallel
b) ![u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5](https://tex.z-dn.net/?f=%20u%20v%3D%20%281%29%2A%282%29%20%2B%20%28-1%29%2A%28-1%29%2B%20%282%29%2A%281%29%3D5)
![|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}](https://tex.z-dn.net/?f=%20%7Cu%7C%3D%20%5Csqrt%7B%281%29%5E2%20%2B%28-1%29%5E2%20%2B%282%29%5E2%7D%3D%5Csqrt%7B6%7D)
![|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}](https://tex.z-dn.net/?f=%20%7Cv%7C%20%3D%5Csqrt%7B%282%29%5E2%20%2B%28-1%29%5E2%20%2B%281%29%5E2%7D%3D%5Csqrt%7B6%7D)
![cos \theta = \frac{uv}{|u| |v|}](https://tex.z-dn.net/?f=%20cos%20%5Ctheta%20%3D%20%5Cfrac%7Buv%7D%7B%7Cu%7C%20%7Cv%7C%7D)
![\theta = cos^{-1} (\frac{uv}{|u| |v|})](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20cos%5E%7B-1%7D%20%28%5Cfrac%7Buv%7D%7B%7Cu%7C%20%7Cv%7C%7D%29)
![\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20cos%5E%7B-1%7D%20%28%5Cfrac%7B5%7D%7B%5Csqrt%7B6%7D%20%5Csqrt%7B6%7D%7D%29%3Dcos%5E%7B-1%7D%20%28%5Cfrac%7B5%7D%7B6%7D%29%20%3D%2033.557)
Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.
c) ![u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0](https://tex.z-dn.net/?f=%20u%20v%3D%20%28a%29%2A%28-b%29%20%2B%20%28b%29%2A%28a%29%2B%20%28c%29%2A%280%29%3D-ab%20%2Bba%20%2B0%20%3D%20-ab%2Bab%20%3D0)
Since the dot product is equal to zero then the two vectors are orthogonal.
Step-by-step explanation:
For each 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.
Part a
u=[-3,9,6], v=[4,-12,-8,]
The dot product on this case is:
![u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168](https://tex.z-dn.net/?f=%20u%20v%3D%20%28-3%29%2A%284%29%20%2B%20%289%29%2A%28-12%29%2B%20%286%29%2A%28-8%29%3D-168)
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:
![|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}](https://tex.z-dn.net/?f=%20%7Cu%7C%3D%20%5Csqrt%7B%28-3%29%5E2%20%2B%289%29%5E2%20%2B%286%29%5E2%7D%3D%5Csqrt%7B126%7D)
![|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}](https://tex.z-dn.net/?f=%20%7Cv%7C%20%3D%5Csqrt%7B%284%29%5E2%20%2B%28-12%29%5E2%20%2B%28-8%29%5E2%7D%3D%5Csqrt%7B224%7D)
And finally we can calculate the angle between the vectors like this:
![cos \theta = \frac{uv}{|u| |v|}](https://tex.z-dn.net/?f=%20cos%20%5Ctheta%20%3D%20%5Cfrac%7Buv%7D%7B%7Cu%7C%20%7Cv%7C%7D)
And the angle is given by:
![\theta = cos^{-1} (\frac{uv}{|u| |v|})](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20cos%5E%7B-1%7D%20%28%5Cfrac%7Buv%7D%7B%7Cu%7C%20%7Cv%7C%7D%29)
If we replace we got:
![\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20cos%5E%7B-1%7D%20%28%5Cfrac%7B-168%7D%7B%5Csqrt%7B126%7D%20%5Csqrt%7B224%7D%7D%29%3Dcos%5E%7B-1%7D%20%28-1%29%20%3D%20%5Cpi)
Since the angle between the two vectors is 180 degrees we can conclude that are parallel
Part b
u=[1,-1,2] v=[2,-1,1]
The dot product on this case is:
![u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5](https://tex.z-dn.net/?f=%20u%20v%3D%20%281%29%2A%282%29%20%2B%20%28-1%29%2A%28-1%29%2B%20%282%29%2A%281%29%3D5)
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:
![|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}](https://tex.z-dn.net/?f=%20%7Cu%7C%3D%20%5Csqrt%7B%281%29%5E2%20%2B%28-1%29%5E2%20%2B%282%29%5E2%7D%3D%5Csqrt%7B6%7D)
![|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}](https://tex.z-dn.net/?f=%20%7Cv%7C%20%3D%5Csqrt%7B%282%29%5E2%20%2B%28-1%29%5E2%20%2B%281%29%5E2%7D%3D%5Csqrt%7B6%7D)
And finally we can calculate the angle between the vectors like this:
![cos \theta = \frac{uv}{|u| |v|}](https://tex.z-dn.net/?f=%20cos%20%5Ctheta%20%3D%20%5Cfrac%7Buv%7D%7B%7Cu%7C%20%7Cv%7C%7D)
And the angle is given by:
![\theta = cos^{-1} (\frac{uv}{|u| |v|})](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20cos%5E%7B-1%7D%20%28%5Cfrac%7Buv%7D%7B%7Cu%7C%20%7Cv%7C%7D%29)
If we replace we got:
![\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20cos%5E%7B-1%7D%20%28%5Cfrac%7B5%7D%7B%5Csqrt%7B6%7D%20%5Csqrt%7B6%7D%7D%29%3Dcos%5E%7B-1%7D%20%28%5Cfrac%7B5%7D%7B6%7D%29%20%3D%2033.557)
Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.
Part c
u=[a,b,c] v=[-b,a,0]
The dot product on this case is:
![u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0](https://tex.z-dn.net/?f=%20u%20v%3D%20%28a%29%2A%28-b%29%20%2B%20%28b%29%2A%28a%29%2B%20%28c%29%2A%280%29%3D-ab%20%2Bba%20%2B0%20%3D%20-ab%2Bab%20%3D0)
Since the dot product is equal to zero then the two vectors are orthogonal.