Question

Which of the following is a unit vector

• A. $(\frac{1}{2},\frac{1}{\sqrt{2}})$
• B. $(\frac{1}{\sqrt{2}},1)$
• C. $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
• D. (1, 1)

Answer 1

The correct option is C $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

Let’s check the magnitude for all these vectors. Whichever vector has the magnitude 1, it will be a unit vector.
so a) $(\frac{1}{2},\frac{1}{\sqrt{2}})$, Magnitude = $\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}=\frac{\sqrt{3}}{2}\ne 1$
b) $(\frac{1}{\sqrt{2}},1)$. Magnitude = $\sqrt{{\left(\frac{1}{2}\right)}^2+(1{)}^2}\ne 1$
c) $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$. Magnitude = $\sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}=1$
So it is a unit vector
d) (1, 1), Magnitude = $\sqrt{(1{)}^2+(1{)}^2}=\sqrt{2}\ne 1$
So only (c) is a unit vector.