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

2 Answers
Apr 12, 2018

sqrt{26}

Explanation:

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}

Apr 12, 2018

AB=sqrt26

Explanation:

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