Respuesta :
Answer:
a) [tex] u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168[/tex]
Since the dot product is not equal to zero then the two vectors are not orthogonal.
[tex] |u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}[/tex]
[tex] |v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}[/tex]
[tex] cos \theta = \frac{uv}{|u| |v|}[/tex]
[tex] \theta = cos^{-1} (\frac{uv}{|u| |v|})[/tex]
If we replace we got:
[tex] \theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi[/tex]
Since the angle between the two vectors is 180 degrees we can conclude that are parallel
b) [tex] u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5[/tex]
[tex] |u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}[/tex]
[tex] |v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}[/tex]
[tex] cos \theta = \frac{uv}{|u| |v|}[/tex]
[tex] \theta = cos^{-1} (\frac{uv}{|u| |v|})[/tex]
[tex] \theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557[/tex]
Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.
c) [tex] u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0[/tex]
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:
[tex] u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168[/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] |u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}[/tex]
[tex] |v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}[/tex]
And finally we can calculate the angle between the vectors like this:
[tex] cos \theta = \frac{uv}{|u| |v|}[/tex]
And the angle is given by:
[tex] \theta = cos^{-1} (\frac{uv}{|u| |v|})[/tex]
If we replace we got:
[tex] \theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi[/tex]
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:
[tex] u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5[/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] |u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}[/tex]
[tex] |v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}[/tex]
And finally we can calculate the angle between the vectors like this:
[tex] cos \theta = \frac{uv}{|u| |v|}[/tex]
And the angle is given by:
[tex] \theta = cos^{-1} (\frac{uv}{|u| |v|})[/tex]
If we replace we got:
[tex] \theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557[/tex]
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:
[tex] u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0[/tex]
Since the dot product is equal to zero then the two vectors are orthogonal.
Answer:
a) parallel
b) neither
c) orthogonal
Step-by-step explanation:
a) [tex]u.v = -3*4-9*12-6*8=-168\neq 0[/tex]
So, they are not orthogonal.
3<-1,3,2> and 4<1,-3,-2>
The angle between them is 180°.
So they are parallel.
b) [tex]u.v=1*2+1*1+2*1=5\neq0[/tex]
So, they are not orthogonal.
<1, -1, 2> and <2, -1, 1>
The angle between them is 180°.
So, they are parallel.
c) [tex]u.v = -a.b+a.b+c.0=0[/tex]
So, they are orthogonal.
