What is the distance between # (–2, 1, 3) # and #(8, 6, 0)
#?
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"How do you find the area of the region bounded by the polar curves #r=1+cos(theta)# and #r=1-cos(theta)# ?"
#"Distance" = 11.6" units to 3 significant figures"#
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)#.
