Question

Determine whether the given vectors are orthogonal, parallel or neither. (a) u=[-3,9,6], v=[4,-12,-8,], (b) u=[1,-1,2] v=[2,-1,1] (c) u=[a,b,c] v=[-b,a,0]

Answer 1

a) parallel

b) neither

c) orthogonal

Explanation:

a)
`u.v = -3*4-9*12-6*8=-168\\eq 0`

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)
`u.v=1*2+1*1+2*1=5\\eq0`

So, they are not orthogonal.

<1, -1, 2> and <2, -1, 1>

The angle between them is 180°.

So, they are parallel.

c)
`u.v = -a.b+a.b+c.0=0`

So, they are orthogonal.

Answer 2

a)
`u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168`

Since the dot product is not equal to zero then the two vectors are not orthogonal.

`|u|= √((-3)^2 +(9)^2 +(6)^2)=√(126)`

`|v| =√((4)^2 +(-12)^2 +(-8)^2)=√(224)`

`cos \theta = (uv)/(|u| |v|)`

`\theta = cos^(-1) ((uv)/(|u| |v|))`

If we replace we got:

`\theta = cos^(-1) ((-168)/(√(126) √(224)))=cos^(-1) (-1) = \pi`

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`

`|u|= √((1)^2 +(-1)^2 +(2)^2)=√(6)`

`|v| =√((2)^2 +(-1)^2 +(1)^2)=√(6)`

`cos \theta = (uv)/(|u| |v|)`

`\theta = cos^(-1) ((uv)/(|u| |v|))`

`\theta = cos^(-1) ((5)/(√(6) √(6)))=cos^(-1) ((5)/(6)) = 33.557`

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`

Since the dot product is equal to zero then the two vectors are orthogonal.

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`

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|= √((-3)^2 +(9)^2 +(6)^2)=√(126)`

`|v| =√((4)^2 +(-12)^2 +(-8)^2)=√(224)`

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

`cos \theta = (uv)/(|u| |v|)`

And the angle is given by:

`\theta = cos^(-1) ((uv)/(|u| |v|))`

If we replace we got:

`\theta = cos^(-1) ((-168)/(√(126) √(224)))=cos^(-1) (-1) = \pi`

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`

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|= √((1)^2 +(-1)^2 +(2)^2)=√(6)`

`|v| =√((2)^2 +(-1)^2 +(1)^2)=√(6)`

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

`cos \theta = (uv)/(|u| |v|)`

And the angle is given by:

`\theta = cos^(-1) ((uv)/(|u| |v|))`

If we replace we got:

`\theta = cos^(-1) ((5)/(√(6) √(6)))=cos^(-1) ((5)/(6)) = 33.557`

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`

Since the dot product is equal to zero then the two vectors are orthogonal.