Solving Vectors: Find Dot & Cross Prods of OX,YX & OY,YX

[Kawakaze]

Homework Statement

I already made a start to this question, the position vectors below are worked out in the first part of it, I am 99% sure they are correct so, on to the bit I am stuck on. :)

A player stands with his feet at the point O and serves the ball from a point Y at
a height of 2.4000m vertically above O. Assume that the ball then travels in a
straight line directly over the net at the midpoint A, which is at a height of
0.9144 m, before bouncing at the point X.

diagram here = https://www.physicsforums.com/attachment.php?attachmentid=32722&d=1299067106

find the dot product of the following position vectors, OX & YX, hence find the angle between them

a=OX=[19.2038i, 6.6475j, 0k]
b=YX=[19.2038i, 6.6475j, -2.4k]

find the cross product of the following position vectors, OY & YX, hence find the area of the triangle they form

a=OY=[0i, 0j, 2.4k]
b=YX=[19.2038i, 6.6475j, -2.4k]

Homework Equations

a(dot)b = |a||b|cos$$\theta$$
a x b = absin$$\theta$$

The Attempt at a Solution

Not sure how to work this with vectors, I've come across vectors before and the dot and cross product before, but never the two together.

Some hints would be great. :)

 

[gneill]

Do the products with their Cartesian components:

$$A \cdot B = A_x B_x + A_y B_y + A_z B_z$$

$$A \times B = \left| \begin{array}{ccc} \hat i \ \ \ \hat j \ \ \ \hat k \ \ \\ A_x \ A_y \ A_z \ \\ B_x \ B_y \ B_z \ \\ \end{array} \right|$$

 

[Kawakaze]

So that would be

19.2038*19.2038i + 6.6475*6.6475j + 0*-2.4k = |a||b|cos$$\theta$$

what are the a and b terms on the RHS related too?

btw thanks for speedy response :)

 

[gneill]

The dot product is a scalar quantity. So drop the i,j,k.
|a| is the magnitude of vector A.
|b| is the magnitude of vector B.
θ is the angle between the vectors.

 

[Kawakaze]

One thing that's troubling me, is the -2.4k term correct? that would be b3 on the cross product part of the question

Ive just run the numbers and for the cross product and i get

a x b = -15.954i + 46.0891j + 0k

How does this apply to the area of a triangle?

 

[gneill]

One thing that's troubling me, is the -2.4k term correct? that would be b3 on the cross product part of the question

"Correct" in what sense? I didn't check your derivation of the vectors. I only offered a way to compute the dot and cross products from them.

If one or more components of the vectors happens to be zero, it doesn't invalidate the methods; They are general.

 

[Kawakaze]

I mean is it ok to have a negative term in these general methods? if you look at my post above, youll see what i got at the end of it. I can work out the area of the triangle the normal way and it doesn't match with this result

 

[gneill]

I mean is it ok to have a negative term in these general methods? if you look at my post above, youll see what i got at the end of it. I can work out the area of the triangle the normal way and it doesn't match with this result

A negative component in the cross product is okay; it's a vector normal to the plane that the two vectors define. What's the magnitude of the result? That should be the area of the parallelogram defined by the two vectors. The area of the triangle would be half of that.

 

[Kawakaze]

I get the magnitude to be 52.417, so the area of the triangle is half of this =26.2085

I tried to check it using 1/2bh, which is 0.5 x 19.2038 x 2.4 = 23.0446

Id expect the area of the triangle to be the same no matter which method I used, so its clear something is wrong here. :)

 

[gneill]

For your vectors

a = [0i, 0j, 2.4k]
b = [19.2038i, 6.6475j, -2.4k]

I get the cross product a x b = [-15.954, 46.089, 0]

It's magnitude is |a x b| = 48.772

So that the area of the triangle is 48.772/2 = 24.386

When I work out the area of the triangle "manually", I get the same number.