What is the distance between # (–2, 1, 3) # and #(8, 6, 0) #?

First, calculate your distance per dimension:

• #x : 8+2 = 10#
• #y : 6-1 = 5#
• #z : 3+-0 = 3#

Next, apply 3D Pythagoras' theorem:

#h^2 = a^2 + b^2 + c^2#

Where:

• #h^2# is the square of the distance between two points
• #a^2#, #b^2#, and #c^2# are the calculated dimensional distances

We can adjust the theorem to solve directly for #h#:

#h = sqrt(a^2 + b^2 + c^2)#

Finally, substitute your values into the equation and solve:

#h = sqrt(10^2 + 5^2 + 3^2)#
#h = sqrt(100 + 25 + 9)#
#h = sqrt(134)#
#h = 11.5758369028 = 11.6" to 3 significant figures"#

#:. "Distance" = 11.6" units to 3 significant figures"#

The distance formula for Cartesian coordinates is

#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2#
Where #x_1, y_1,z_1#, and#x_2, y_2,z_2# are the Cartesian coordinates of two points respectively.
Let #(x_1,y_1,z_1)# represent #(-2,1,3)# and #(x_2,y_2,z_2)# represent #(8,6,0)#.
#implies d=sqrt((8-(-2))^2+(6-1)^2+(0-3)^2#
#implies d=sqrt((10)^2+(5)^2+(-3)^2#
#implies d=sqrt(100+25+9#
#implies d=sqrt(134#

Hence the distance between the given points is #sqrt(134)#.