Cross Product Vectors: Finding Direction and Magnitude

[popo902]

Homework Statement

You are looking down at a map. A vector u with ||u||
= 5 points north and a vector v with ||v|| = 10 points northeast.
The cross product u×v points:
A) south
B) northwest
C) up
D) down
Please enter the letter of the correct answer:
The magnitude ||u×v|| =

Homework Equations

The Attempt at a Solution

it says that the correct answer is "D"
why does it point down?
it says the correct answer (for the magnitude) is (5*10*sqrt(2) )/2
but i don't even know how to get that.
Can someone explain how I'm suppose to get that?

 

[Office_Shredder]

Do you know what the formula is for the magnitude of the cross product of two vectors?

 

[LCKurtz]

Homework Statement

You are looking down at a map. A vector u with ||u||
= 5 points north and a vector v with ||v|| = 10 points northeast.
The cross product u×v points:
A) south
B) northwest
C) up
D) down
Please enter the letter of the correct answer:
The magnitude ||u×v|| =

Homework Equations

The Attempt at a Solution

it says that the correct answer is "D"
why does it point down?
it says the correct answer (for the magnitude) is (5*10*sqrt(2) )/2
but i don't even know how to get that.
Can someone explain how I'm suppose to get that?

1. Do you know how to write the two vectors in 3D with i, j, and k?
2. Do you know the algebraic formula for calculating a cross product?
3. Do you know the geometric properties of the cross product (right hand rule)?

Assuming (2) is yes, to see it points down you could calculate the cross product and look at the answer.
Assuming (3) is yes, check out the right hand rule for your two vectors.

If you can't answer 1,2, and 3, yes, hit the book some more and read about it.

 

[popo902]

oh, okay okay
the formula for the magnitude of cros product of two vectors was under the torque section
its mag(a) * mag(b) (sin //theta) right?
now i see, the angle is 45 degrees because it's bisects the north and the east vecors
so it really is 5 * 10 * (1/sqrt(2))
i'm still working on understanding the right hand rule, but I'm sure i'll get it
thanks for the clarification tho

 

[D H]

Orient your hand so your thumb points along u, i.e., to the north. Point your index finger along v: northeast. Where is your middle finger pointing?

 

[Office_Shredder]

Orient your hand so your thumb points along u, i.e., to the north. Point your index finger along v: northeast. Where is your middle finger pointing?

What? Mine's pointing the same way as my index finger. I guess the point is I'm supposed to flip you the bird? I think this test is a bit ambiguous

 

[D H]

Rather than a smart-mouth reply, why don't you try helping?

Forming a right-hand set of axes: Make each pair of (thumb, index finger), (thumb, middle finger), and (index finger, middle finger) form a right angle. Designate x as the direction pointed by your thumb, y by your index finger, z by your middle finger. Then (x,y,z) forms a right-handed coordinate system.

Cross product: The determine the direction of \mathbf u \times \mathbf v, point your thumb in the direction of \mathbf u, your index finger in the direction of \mathbf v. Point your middle finger so it is perpendicular to both your thumb and index finger. Your middle finger now points in the direction of \mathbf u \times \mathbf v.

 

[LCKurtz]

oh, okay okay
the formula for the magnitude of cros product of two vectors was under the torque section
its mag(a) * mag(b) (sin //theta) right?
now i see, the angle is 45 degrees because it's bisects the north and the east vecors
so it really is 5 * 10 * (1/sqrt(2))
i'm still working on understanding the right hand rule, but I'm sure i'll get it
thanks for the clarification tho

I asked you three questions, which weren't intended to be rhetorical:

1. Do you know how to write the two vectors in 3D with i, j, and k?
2. Do you know the algebraic formula for calculating a cross product?
3. Do you know the geometric properties of the cross product (right hand rule)?

Have you made progress on answering them? That will solve your problem.

 

[rarkup]

Think of the map with the vectors drawn on as the xy-plane. The z-axis points directly up out of the map, and is positive in the upward direction.

North would be directly along the y-axis, and north-east is 45 degrees between the x and y axis.

The cross-product either points up out of the map along the positive z axis, or in the opposite direction, so all you need to do is determine which. You can therefore make your job easier by scaling the 2 vectors down to sizes that are easier to work with - namely, divide both vector lengths by 5, giving ||u|| = 1 and ||v|| =2. The resulting cross-product of these two scaled down vectors will have a different length than that of your original 2 vectors, but it will point in the same direction, and that's all you need to find out.

As u has length 1, it can be represented by the vector (0,1,0).
v has length 2, but since it is the hypotenuse of a 45 degree right angled-triangle, the x an y components of v must both be sqrt(2). So v is the vector (\sqrt{2}, \sqrt{2}, 0).

Doing the cross-product of these two shorter vectors tells you whether it goes up (positive) or down (negative).

 

[popo902]

I asked you three questions, which weren't intended to be rhetorical:

1. Do you know how to write the two vectors in 3D with i, j, and k?
2. Do you know the algebraic formula for calculating a cross product?
3. Do you know the geometric properties of the cross product (right hand rule)?

Have you made progress on answering them? That will solve your problem.

i can write the vectro with form i,j,k
i can formulate a cross product of two vectors (but don't you need x,y,z coordinates to find the detrminate and get the product?)
now i see the concept of the right hand rule...i was just making a thiumbs up sign, but i was supposed to aligning the three fingers seprately

so North vector is my thum
the northest vector would be my mindex, but pointing east
and my middle is the cross product of the two vectors which is orthogonal to both
so i point my index east ,my hand flips aroung and the cross prod vector points down, right?

 

[LCKurtz]

Yes, the right hand rule shows it points down. Since your two vectors lie in the xy plane, they have a z component of 0.

u = 0i + 5j + 0k
v = 10 cos(Pi/4)i + 10 sin(Pi/4)j + 0k
= 10/sqrt(2) i + 10/sqrt(2) j + 0 k

Try calculating u X v and see if it doesn't come out pointing down. That's how you can do it analytically.