Question

Determine all the real numbers for which the vector <1-c,c/3> is a unit vector.

Answer 1

The vector is a unit vector if its magnitude is 1:
\[\sqrt{(1-c)^2+\left(\frac{c}{3}\right)^2}=1\]

Answer 2

yea so i was thinking then it could be everything except 0 and 1 , also mayb am overthinking it

Answer 3

Actually if \(c=0\), the magnitude is 1, making it a unit vector for this \(c\).

Answer 4

lol am really over thinking it lmao

Answer 5

All you have to do is solve for \(c\):
\[\sqrt{(1-c)^2+\left(\frac{c}{3}\right)^2}=1\]

Answer 6

so it would jus be 0 ?

Answer 7

Well, let's see:
\[(1-c)^2+\left(\frac{c}{3}\right)^2=1\]
\[1-2c+c^2+\frac{c^2}{9}=1\]
\[9-18c+9c^2+c^2=9\]
\[10c^2-18c=0\]
\[2c\left(5c-9\right)=0\]
\[c=0~~\text{and}~~c=\frac{9}{5}\]

Answer 8

i see, i guess am just getting mixed up with the formula of u/|u|

Answer 9

Ah, that's the unit vector in the *direction* of u. All you need here is the magnitude

Answer 10

oooohh

Answer 11

thx