Find Distance in R^3

by **nycmath** » Wed Sep 02, 2026 5:01 am

See attachment.

by **Math Tutor** » Wed Sep 02, 2026 12:11 pm

[tex]d=\sqrt{(4-1)^2+(6-2)^2+\bigl(7-(-5)\bigr)^2}=\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13[/tex]

by **nycmath** » Thu Sep 03, 2026 3:58 am

Math Tutor wrote:[tex]d=\sqrt{(4-1)^2+(6-2)^2+\bigl(7-(-5)\bigr)^2}=\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13[/tex]

One of the easiest calculations in Calculus lll and fun, too.

by **Eigenvalue** » Thu Sep 03, 2026 11:27 pm

Here is my work for the question:

by **nycmath** » Sat Sep 05, 2026 10:29 am

Eigenvalue wrote:Here is my work for the question:

Very good, honestly, but a written explanation of your work is needed. Do you agree?