What is the distance between #(4, 1, –3) # and #(0, 4, –2) #?

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Answer 1 · 2018-04-12T08:49:43

#sqrt{26}#

The distance is equal to the magnitude of the vector between the two points which can be expressed as: #|((4), (1),( -3)) - ((0),(4),(-2))|#

#|((4 -0), (1-4), (-3-(-2)))|#

#|((4), (-3), (-1))|#

The magnitude is #sqrt{(4)^2 + (-3)^2 + (-1)^2}#

#sqrt{16 + 9 + 1}# = #sqrt{26}#

Answer 2 · 2018-04-12T09:21:57

#AB=sqrt26#

We know that;

If #AinRR^3 and BinRR^3#,then the distance between

#A(x_1,y_1,z_1) and B(x_2,y_2,z_2) # is

#AB=|vec(AB)|=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#

Where, #vec(AB)=(x_2-x_1,y_2-y_1,z_2-z_1)#

We have, #A(4,1,-3) and B(0,4,-2)#

#=>AB=sqrt((4-0)^2+(1-4)^2+(-3+2)^2)#

#=>AB=sqrt(16+9+1#

#=>AB=sqrt26#