Question

A particle has a position vector \[(3\widehat i + 2\widehat k)m\] at time t=0. It moves with constant velocity$( - \widehat i + 3\widehat k)m{s^{ - 1}}$ The position vector of the particle after 3s is \[(x\widehat i + \widehat {vj} + z\widehat k)\]. Then find $x + y - z$

Answer 1

Hint: Displacement and velocity in two or three dimensions are direct extensions of the one-dimensional definitions. Since they are vector quantities the calculations have to be done by the rules of the vector.


Complete step by step solution:
Position Vector: The position vector of an object at a certain time is the position of the object w.r.t to the origin. It is denoted by a straight line between the origin and the position at the time.
Displacement Vector: The displacement vector of an object between two points is the straight line between the two points regardless of the path followed. The path length is always equal to or more than the displacement.
The change in the position vector of an object can be called the displacement vector.
The position vector is used to indicate the position of a certain body. If we know the position of a body is important when it comes to describing the motion of that body. The position vector of an object is measured directly from the origin, in general.
Position vector \[\overrightarrow {{r_{}}} = (\widehat i + \widehat j + \widehat k)\]
Where,
\[\widehat i = \]Unit vector along the x-direction
\[\widehat j = \] Unit vector along the y-direction
\[\widehat k = \]Unit vector along the z-direction
From the final equation for $x = 0$
And we take $y = 2$
Then $z = - 7$
Therefore we get $x + y - z = 0 + 2 - ( - 7) = 9$
Then we take position vector \[\overrightarrow {{r_n}} = (3\widehat i - \widehat j + 2\widehat k)m\]
And the constant velocity is given by the equation $\overrightarrow v = ( - \widehat i + \widehat j + 3\widehat k)m{s^{ - 1}}$
Then the position vector at $\overrightarrow r (t = 3) = \,\,\overrightarrow v \,.\,t\, + \overrightarrow {{r_n}} $
We get the equation as
$\overrightarrow r (t = 3) = \,\,( - \widehat i + \widehat j + 3\widehat k).\,t\, + (3\widehat i - \widehat j + 2\widehat k)$
$\overrightarrow r (t = 3) = - 3\widehat i + 3\widehat j - 9\widehat k + 3\widehat i - \widehat j + 2\widehat k$
The position vector at $\overrightarrow r (t = 3) = (0\widehat i + 2\widehat j - \widehat 7k)m$

Note: The component of the displacement vector can either be positive when the final position is larger than the initial one. It can also be negative if the final position is smaller than the initial one. It can then be zero if the object ends at the starting point. While plotting the displacement it gives us the information and meaning to the unit vector solution to the problem. When plotting the displacement, we need to make sure that we include its components as well as its magnitude and the angle that it makes with a chosen axis.