What is the speed of an object that travels from #( 4,-2,2) # to #( -3, 8,-7 ) # over #3 s#?

1 Answer
Apr 4, 2018

Answer:

The answer would be the distance between the two points (or vectors) divided by the time. So you should get #(sqrt(230))/3# units per second.

Explanation:

To get the distance between the two points (or vectors), just use the distance formula #d = sqrt(x^2 +y^2 +z^2)# on the difference between the two given points.

ie #(x,y,z) = (-3-4, 8 -(-2),-7-2) = (-7,10,-9)# ( note : it does not matter which way around we substract the points since the formula uses squares and thus eliminates any negative signs. We can do point A - point B or point B - point A)

Now applying the distance formula, we get
#d = sqrt((-7)^2 +(10)^2 +(-9)^2) = sqrt(230)#

Then all that is left is to divide by the time to get the answer.

Interesting fact: This distance formula is actually called the Euclidean Norm in the real normed space #R^n#, denoted by #||\bar(x)||_2#.