Let $MNOP$ be a square with vertices in order. If $t$ and $f$ are the magnitudes of the vectors $\overrightarrow{M}\overrightarrow{N}$ and $\overrightarrow{N}\overrightarrow{O}$ respectively, what are the values of $t$ and $f$ if the side length of the square is $1$? $t$ = $f$ =

• A. $t=1,f=1$
• B. $t=\sqrt{2},f=\sqrt{2}$
• C. $t=0,f=0$
• D. $t=2,f=2$

Answer: (A)

Verified step by step guidance

1. Identify the vectors involved: \( \overrightarrow{MN} \) is the vector from vertex M to vertex N, and \( \overrightarrow{NO} \) is the vector from vertex N to vertex O of the square MNOP.
2. Recall that in a square, all sides have equal length. Since the side length is given as 1, the magnitude of each side vector should be 1.
3. The magnitude of a vector \( \overrightarrow{AB} = (x, y) \) is calculated using the formula: \( |\overrightarrow{AB}| = \sqrt{x^2 + y^2} \).
4. Since \( \overrightarrow{MN} \) and \( \overrightarrow{NO} \) represent adjacent sides of the square, their magnitudes correspond to the side length, which is 1.
5. Therefore, both \( t = |\overrightarrow{MN}| \) and \( f = |\overrightarrow{NO}| \) equal the side length of the square, which is 1.