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

2 Answers
Jan 30, 2016

The distance between these points is given by r=sqrt((0-(-4))^2+(4-0)^2+((-2)-2)^2) and is 4sqrt3 or 6.93 units.

Explanation:

The distance, r, between two points in 3 dimensions is given by:

r=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)

Substituting in the coordinates for the two points given:

r=sqrt((0-(-4))^2+(4-0)^2+((-2)-2)^2)
= sqrt((-4)^2+(4)^2+(-4)^2)
= sqrt(16+16+16) = sqrt48 = 4sqrt3 = 6.93

Jan 30, 2016

6.928

Explanation:

suppose,
x_1=-4

y_1=0

z_1=2

x_2=0

y_2=4

z_2=-2

now, if we find out the position vector of the two points for main point O(0,0,0), we get,
vec(OA)=-4i+2k

vec(OB)=4j-2k

we know,

vec(AB)=vec(OB)-vec(OA)

=(4j-2k)-(4i+2k)

=-4i+4j-2k-2k

=-4i+4j-4k

so, the distanse is,

|vec(AB)|=sqrt((-4)^2+4^2+(-4)^2)

=sqrt(48)

=6.928