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

2 Answers
Apr 3, 2018

Distance b/w the pts.=sqrt5 units.

Explanation:

let the pts. be A(3,-1,1) & B(2,-3,1)
so, By distance formula
AB=sqrt(((x_2-x_1)^2)+(y_2-y_1)^2+(z_2-z_1)^2)
AB=sqrt[(2-3)^2+(-3+1)^2+(1-1)^2]
AB=sqrt[1+4+0]
AB=sqrt5 units.

Apr 3, 2018

The distance between (3,-1,1) and (2,-3,1) is sqrt(5)~~2.236.

Explanation:

If you have a point (x_1,y_1,z_1) and another point (x_2,y_2,z_2) and you want to know the distance, you can use the distance formula for a normal pair of (x,y) points and add a z component. The normal formula is d=sqrt((x_2-x_1)^2+(y_2-y_1)^2), so when you add a z component, it becomes d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2). For your points, you would say sqrt((2-3)^2+((-3)-(-1))^2+(1-1)^2) which simplifies to sqrt(5) ~~2.236