Solving Vector Cross Product Homework

[oreosama]

Homework Statement

vector a = (4i+3j-2k)
vector b = (2i-3j+2k)

1. a x b
2. 3a x 2b
3. |3a x 2b|

The Attempt at a Solution

1. -12j - 18k

2. 6(a x b)
6(-12j - 18k)
-48j -108k

3. |6(a x b)|
sqrt(48^2 + 108^2)

sqrt(13968)


idk if I am using identities wrong or if I am way off, thanks for confirmation on right/wrong and any help

 

[ehild]

Homework Statement

vector a = (4i+3j-2k)
vector b = (2i-3j+2k)

1. a x b
2. 3a x 2b
3. |3a x 2b|

The Attempt at a Solution

1. -12j - 18k

2. 6(a x b)
6(-12j - 18k)
-48j -108k

3. |6(a x b)|
sqrt(48^2 + 108^2)

sqrt(13968)idk if I am using identities wrong or if I am way off, thanks for confirmation on right/wrong and any help

Hi oreosama, welcome to PF. Your work is correct except a siliy mistake, How much is 12*6?

ehild

 

[oreosama]

thanks for that.

using the same vectors

a=(4i+3j-2k)
b=(2i-3j+2k)

the vectors a and b define a plane surface. determine a possible vector perp. to that surface. determine a possible vector parallel to that surface.

perp: a x b= -12j -18k

assuming that's right..

parralel: I have no idea :(

i know parallel vectors just have scaler multiplier but I am working with 2 vectors(of which I am assuming intersect to create a plane?). I am pretty confused at this point so I am going to sleep and hope someone provides insight by the time i wake up. thanks for any help

 

[CAF123]

If <a> and <b> are contained within the plane, then a vector parallel to <a> or <b> will also be parallel to the plane.
If two vectors are parallel, then their cross product is the zero vector. Therefore compute:
$$\vec{a}\, ×\,\vec{c} = 0$$ where $\vec{c}$ is the vector you want. You can solve a linear system of 3 variables, and get a dependence on the components of <c>.
Note: <b> can be used in place of <a>

 

[ehild]

The vectors a and b lay in the surface, and so do their linear combinations. All are parallel with the plane. (you can shift a vector parallel with itself, it is the same vector.)

ehild