Vector Distance

Question

Find the distance from (-2,9,-12) to the x-axis.

If you could explain how you get to your answer, I would greatly appreciate it!

Answer 1

(-2,9) is a distance of 9 to the x axis. the x axis is the line y=0

(x,y) = (-2,9) y=9. from y=9 to y=0 is a distance of 9-0 = 9

same with 3 dimenstions

(-2,9,-12) = (x,y,z) with y=9.

y=9 is 9 units from y=0 which is the x axis

(-2,0,-12) is on the y axis and the closest point on y=0 from (-2,9,-12)

(-2.0,-12) to (-2,9,-12) is a distance of 9

d^2 = (-2+2)^2 + 9^2 +(-12+12)^= 0+81+0

d = sqr81 = 9