Calculus Problem

[andrey]

Can anyone give me a hint here...

1. If v cross w = 2i-3j+5k, and v dot w = 3, find tangent Q, where Q is the angle between vectors v and w

2. Find vector which is parallel to the intersection of two planes: 2x-3y+5z=2 and 4x+y-3z=7

 

[SuperGroove]

Tangent = derivative. Me thinks.

VXW = 3

so

3 = 2i-3J+5K

The derivative of that is:
0=2-3+5
0=-4

Okay so that's not right. Think of it as a bump. a Stupid bump. Man...I'm so happy i just PASSED calculus...but I wish I had gotten a better grade.

 

[bleeb]

I think you should write that out again.. do you mean V cross W?

 

[Woodchuck2000]

I'm not sure what you mean by 1) but 2) is quite easy.
To find a vector parallel to the line of intersection, find the normals of both planes and then the vector product of the normals. This is parallel to the intersection.

 

[andrey]

Woodchuck2000, thank you for #2!

As far as number 1, I meant if v cross w = 2i-3j+5k, and v cross w = 3, find tangent Q, where Q is the angle between vectors v and w.

Thanks again,
-- Andrey

 

[Rob9874]

42














just kidding, I don't know.

 

[andrey]

does anybody have an idea for #1 ?

 

[Oscar1613]

do you mean v dot w =3?😕

 

[agnitrate]

Originally posted by: andrey
Can anyone give me a hint here...

1. If v cross w = 2i-3j+5k, and v cross w = 3, find tangent Q, where Q is the angle between vectors v and w

2. Find vector which is parallel to the intersection of two planes: 2x-3y+5z=2 and 4x+y-3z=7

Ok. The angle between V and W is pretty simple.
the formula for it is (where x = angle between the vectors)

cos x = u . v / ( mag (u) * mag(v) )

Ok, so that looks weird. It's the dot product of u and v divided by their individual lengths.

To find a vector parallel to the intersection of two planes you take the cross product of each of their normals. Think that if the planes intersect, it's the same as the intersection between their two normal vectors. Then if you find the cross-product of those two normals, you find a vector that is perpendicular to both of their normals and therefore is parallel to their intersection.

Got it?

-silver

 

[TuxDave]

Originally posted by: andrey
Can anyone give me a hint here...

1. If v cross w = 2i-3j+5k, and v cross w = 3, find tangent Q, where Q is the angle between vectors v and w

v cross w = 3? I thought cross products resulted in vectors, not magnitudes?

 

[dighn]

Originally posted by: andrey
Can anyone give me a hint here...

1. If v cross w = 2i-3j+5k, and v cross w = 3, find tangent Q, where Q is the angle between vectors v and w

2. Find vector which is parallel to the intersection of two planes: 2x-3y+5z=2 and 4x+y-3z=7

1.

i assume u mean v dot w = 3 b/c what u said makes no sense at all 😀

magnitude wise:

cross: vwsin(Q)=mag(result)=sqrt(4+9+25) = sqrt(38)
dot: vwcos(Q) = 3

divide the top two lines you get
sin(q)/cos(q) = sqrt(38)/3
but sin(q)/cos(q) = tan(q)

u have ur answer

2.

whatever the intersection is, the vector must be on both planes ie must be normal to the plane normal vectors
the plane normal vectors are (2,-3,5) and (4,1,-3)

cross (2,-3,5) and (4,1,-3) and u have ur answer

 

[andrey]

Originally posted by: dighn

Originally posted by: andrey
Can anyone give me a hint here...

1. If v cross w = 2i-3j+5k, and v cross w = 3, find tangent Q, where Q is the angle between vectors v and w

2. Find vector which is parallel to the intersection of two planes: 2x-3y+5z=2 and 4x+y-3z=7

1.

i assume u mean v dot w = 3 b/c what u said makes no sense at all 😀

magnitude wise:

cross: vwsin(Q)=mag(result)=sqrt(4+9+25) = sqrt(38)
dot: vwcos(Q) = 3

divide the top two lines you get
sin(q)/cos(q) = sqrt(38)/3
but sin(q)/cos(q) = tan(q)

u have ur answer

2.

whatever the intersection is, the vector must be on both planes ie must be normal to the plane normal vectors
the plane normal vectors are (2,-3,5) and (4,1,-3)

cross (2,-3,5) and (4,1,-3) and u have ur answer

dighn, thank you so much!!!!

And you're correct, I meant dot in nimber 1. It was pretty late at night and I was trying to translate the whole thing from Russian 🙂

Thanks again,
-- Andrey