How to find a vector A that has the same direction as ⟨−8,7,8⟩ but has length 3 ?

2 Answers
Zor Shekhtman
Oct 7, 2015

#(-24/sqrt(177),21/sqrt(177),24/sqrt(177))#

Explanation:

The idea is based on a concept of scaling and similarity.
Any vector that "has the same direction" as #(-8,7,8)# has all the coordinates proportional to this given vector and, therefore, can be described by coordinates #(-8f,7f,8f)# where #f# is a scaling factor.

All we need now is to find a scaling factor that leads to a vector with the length #3#.

The length of a vector with coordinates #(-8f,7f,8f)# equals to
#sqrt(64f^2+49f^2+64f^2) = f*sqrt(177)#

So, if we want the length to be equal to #3#, we should choose
#f = 3/sqrt(177)#

The coordinates of a vector with the same direction as #(-8,7,8)# but with the length #3# will be
#(-24/sqrt(177),21/sqrt(177),24/sqrt(177))#

Alan P.
Oct 7, 2015

#barA_3 ~= < -1.80, 1.578, 1.80 >#

Explanation:

Let
#color(white)("XXX")barA_o#: be the original vector #< -8, 7, 8>#
#color(white)("XXX")barA_1#: be unit vector in same direction as #barA_o#
#color(white)("XXX")barA_3#: be vector in same direction as #barA_o# with length of #3#

#|barA_o| = sqrt((-8)^2+7^2+8^2) = sqrt(177) ~= 13.304#

#barA_1 = (barA_o)/|barA_o| = < (-8)/sqrt(177), 7/sqrt(177), 8/sqrt(177) >#

#barA_3 = 3*barA_1#
#color(white)("XXX")= < (-24)/sqrt(177), 21/sqrt(177), 24/sqrt(177) >#

#color(white)("XXX")~= < -1.80, 1.578, 1.80 >#

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