Question

A particle starts at the point P=(−1,1,3)P=(−1,1,3) and moves along a straight line toward Q=(−4,4,5)Q=(−4,4,5) at a speed of 7 cm/sec. Let xx, yy, zz be measured in centimeters and tt be measured in seconds.Find a parametric vector equation for the position of the object.

Answer 1

Answer: (-1,1,-3) + 7 x t x (3,-2,-8)/$\sqrt{77}$

Step-by-step explanation:

The parametric vector equation for the position of particle is given as:

We have P= (-1,1-3)

Q= (-4,4,5)

So PQ vector= (-1+4,1-3, -3-5)

PQ = (3,-2,-8)

Magnitude of this vector= $\sqrt{(3)^2+(-2)^2+(-8)^2}$

Magnitude= $\sqrt{77}$

So unit vector= (3, -2, -8)/$\sqrt{77}$

Now the equation is as:

P + 7 x t x unit vector = (-1,1,-3) + 7 x t x (3,-2,-8)/$\sqrt{77}$