# Help with Kinematics

1. Jul 26, 2009

### harini_5

An object is moving in 3D space with position vector vectot r=(x)i+y(j)+z(k) and velocity vector v=a(i)+b(j)+c(k)
i,j,k are unit vectors along x,y,z axes
If the object is always moving towards origin,
then
(a) a/x =b/y = c/z
(b) (a/x)(b/y)(c/z)<0
(c)ax+by+cz<0
(d)ab/xy >0
x,y,z,a,b,c are functions of time
question may have one or more correct answer(s)
MY APPROACH
integrating velocity eqn,
r=at(i)+bt(y)+ct(z)
which implies,
x=at
y=bt
z=ct
I dont get the correct ans if I proceed this way
wats wrong?

2. Jul 26, 2009

### rl.bhat

Re: kinematics

Angle between position vector and velocity vector is π.
So the dot product of r and v i.e. r.v = - ( xa + yb + zc) < 0

3. Jul 26, 2009

### harini_5

Re: kinematics

plz explain me,how the angle is pi
what does the statement the object is always moving towards origin
imply?

4. Jul 26, 2009

### RoyalCat

Re: kinematics

For the sake of argument, let's simplify it to a one dimensional problem.
The x axis is positive to the right of the origin, and negative to the left of the origin.
The object starts with an initial displacement $$x_0 > 0$$
If the object is to move towards the origin, is its velocity negative or positive?

The dot product is the scalar product of two vectors.

$$\vec i\cdot \vec j \equiv |i|*|j|*\cos{\theta}$$ where $$\theta$$ is the angle between the two vectors.

So for your velocity and displacement vectors, make an analogy from the one-dimensional case.

The velocity runs opposite the displacement vector for each of the axes. Can you see why the dot product is necessarily negative, now?

Last edited: Jul 26, 2009