Question

(1 point) A particle starts at the point P=(−2,−1,3)P=(−2,−1,3) when t=0t=0 and moves along a straight line toward Q=(−4,0,6)Q=(−4,0,6) at a speed of 77 cm/sec. Let x, y, and z be measured in cm, and t in seconds. Find a parametric vector equation for the position of the object. r⃗ (t)=r→(t)=

Answer 1

Answer:

The position is given by r(t) =  (-2 -t √14, 1 + √3.5, 3 + 3√3.5)

Step-by-step explanation:

The direction is given by the difference between the starting point and the end point, thus it is (-4,0,6)-(-2,-1,3) = (-2,1,3). The norm of this vector is        √( (-2)²+1²+3²) = √14 = 3.7416 secons

Since the speed is 7 cm/sec, we need to multiply this vector by 7/√14 = √3.5 in order to obtain how much does the particle advance each second.

As a consecuence, it advances (-2,1,3) * √3.5 = (- √14, √3.5, 3 √3.5)

And, as a result, the position after t seconds is given by the function

r(t) = (-2,1,3) + t (-√14, √3.5, 3 √3.5) = (-2 -t √14, 1 + √3.5, 3 + 3√3.5)