What is the distance between #(–4, 0, 2) # and #(0, 4, –2) #?

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Answer 1 · 2016-01-30T08:00:57

The distance between these points is given by #r=sqrt((0-(-4))^2+(4-0)^2+((-2)-2)^2)# and is #4sqrt3# or #6.93# units.

The distance, #r#, between two points in 3 dimensions is given by:

#r=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#

Substituting in the coordinates for the two points given:

#r=sqrt((0-(-4))^2+(4-0)^2+((-2)-2)^2)#
= #sqrt((-4)^2+(4)^2+(-4)^2)#
= #sqrt(16+16+16) = sqrt48 = 4sqrt3 = 6.93#

Answer 2 · 2016-01-30T08:03:41

6.928

suppose,
#x_1=-4#

#y_1=0#

#z_1=2#

#x_2=0#

#y_2=4#

#z_2=-2#

now, if we find out the position vector of the two points for main point O(0,0,0), we get,
#vec(OA)=-4i+2k#

#vec(OB)=4j-2k#

we know,

#vec(AB)=vec(OB)-vec(OA)#

#=(4j-2k)-(4i+2k)#

#=-4i+4j-2k-2k#

#=-4i+4j-4k#

so, the distanse is,

#|vec(AB)|=sqrt((-4)^2+4^2+(-4)^2)#

#=sqrt(48)#

#=6.928#