Answer :
To determine which vectors are unit vectors, we must confirm whether each vector has a magnitude of exactly 1. The magnitude of a vector [tex]\( \mathbf{v} = \langle v_1, v_2 \rangle \)[/tex] is calculated using the formula:
[tex]\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} \][/tex]
We apply this process to each vector provided:
1. First vector:
[tex]\[ u = \left\langle -\frac{1}{3} \sqrt{\frac{7}{2}}, \frac{1}{3} \sqrt{\frac{13}{2}} \right\rangle \][/tex]
- Calculate each component squared:
[tex]\[ \left(-\frac{1}{3} \sqrt{\frac{7}{2}}\right)^2 = \frac{1}{9} \left(\frac{7}{2}\right) = \frac{7}{18} \][/tex]
[tex]\[ \left(\frac{1}{3} \sqrt{\frac{13}{2}}\right)^2 = \frac{1}{9} \left(\frac{13}{2}\right) = \frac{13}{18} \][/tex]
- Sum these squares and find the square root:
[tex]\[ \frac{7}{18} + \frac{13}{18} = \frac{20}{18} = \frac{10}{9} \][/tex]
[tex]\[ \sqrt{\frac{10}{9}} = \frac{\sqrt{10}}{3} \][/tex]
- Since [tex]\(\frac{\sqrt{10}}{3} \neq 1\)[/tex], this vector is not a unit vector.
2. Second vector:
[tex]\[ u = \left\langle \frac{1}{\sqrt{3}}, -\sqrt{\frac{2}{3}} \right\rangle \][/tex]
- Calculate each component squared:
[tex]\[ \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \][/tex]
[tex]\[ \left(-\sqrt{\frac{2}{3}}\right)^2 = \frac{2}{3} \][/tex]
- Sum these squares and find the square root:
[tex]\[ \frac{1}{3} + \frac{2}{3} = 1 \][/tex]
[tex]\[ \sqrt{1} = 1 \][/tex]
- Since the magnitude is 1, this vector is a unit vector.
3. Third vector:
[tex]\[ u = \left\langle \frac{1}{2} \sqrt{\frac{5}{3}}, -\frac{1}{2} \sqrt{\frac{7}{3}} \right\rangle \][/tex]
- Calculate each component squared:
[tex]\[ \left(\frac{1}{2} \sqrt{\frac{5}{3}}\right)^2 = \frac{1}{4} \left(\frac{5}{3}\right) = \frac{5}{12} \][/tex]
[tex]\[ \left(-\frac{1}{2} \sqrt{\frac{7}{3}}\right)^2 = \frac{1}{4} \left(\frac{7}{3}\right) = \frac{7}{12} \][/tex]
- Sum these squares and find the square root:
[tex]\[ \frac{5}{12} + \frac{7}{12} = 1 \][/tex]
[tex]\[ \sqrt{1} = 1 \][/tex]
- Since the magnitude is 1, this vector is a unit vector.
4. Fourth vector:
[tex]\[ u = \left\langle \frac{3}{\sqrt{7}}, -\frac{2}{\sqrt{7}} \right\rangle \][/tex]
- Calculate each component squared:
[tex]\[ \left(\frac{3}{\sqrt{7}}\right)^2 = \frac{9}{7} \][/tex]
[tex]\[ \left(-\frac{2}{\sqrt{7}}\right)^2 = \frac{4}{7} \][/tex]
- Sum these squares and find the square root:
[tex]\[ \frac{9}{7} + \frac{4}{7} = \frac{13}{7} \][/tex]
[tex]\[ \sqrt{\frac{13}{7}} \neq 1 \][/tex]
- Since the magnitude is not 1, this vector is not a unit vector.
In conclusion, the correct answers are:
[tex]\[ \boxed{2 \text{ and } 3} \][/tex]
[tex]\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} \][/tex]
We apply this process to each vector provided:
1. First vector:
[tex]\[ u = \left\langle -\frac{1}{3} \sqrt{\frac{7}{2}}, \frac{1}{3} \sqrt{\frac{13}{2}} \right\rangle \][/tex]
- Calculate each component squared:
[tex]\[ \left(-\frac{1}{3} \sqrt{\frac{7}{2}}\right)^2 = \frac{1}{9} \left(\frac{7}{2}\right) = \frac{7}{18} \][/tex]
[tex]\[ \left(\frac{1}{3} \sqrt{\frac{13}{2}}\right)^2 = \frac{1}{9} \left(\frac{13}{2}\right) = \frac{13}{18} \][/tex]
- Sum these squares and find the square root:
[tex]\[ \frac{7}{18} + \frac{13}{18} = \frac{20}{18} = \frac{10}{9} \][/tex]
[tex]\[ \sqrt{\frac{10}{9}} = \frac{\sqrt{10}}{3} \][/tex]
- Since [tex]\(\frac{\sqrt{10}}{3} \neq 1\)[/tex], this vector is not a unit vector.
2. Second vector:
[tex]\[ u = \left\langle \frac{1}{\sqrt{3}}, -\sqrt{\frac{2}{3}} \right\rangle \][/tex]
- Calculate each component squared:
[tex]\[ \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \][/tex]
[tex]\[ \left(-\sqrt{\frac{2}{3}}\right)^2 = \frac{2}{3} \][/tex]
- Sum these squares and find the square root:
[tex]\[ \frac{1}{3} + \frac{2}{3} = 1 \][/tex]
[tex]\[ \sqrt{1} = 1 \][/tex]
- Since the magnitude is 1, this vector is a unit vector.
3. Third vector:
[tex]\[ u = \left\langle \frac{1}{2} \sqrt{\frac{5}{3}}, -\frac{1}{2} \sqrt{\frac{7}{3}} \right\rangle \][/tex]
- Calculate each component squared:
[tex]\[ \left(\frac{1}{2} \sqrt{\frac{5}{3}}\right)^2 = \frac{1}{4} \left(\frac{5}{3}\right) = \frac{5}{12} \][/tex]
[tex]\[ \left(-\frac{1}{2} \sqrt{\frac{7}{3}}\right)^2 = \frac{1}{4} \left(\frac{7}{3}\right) = \frac{7}{12} \][/tex]
- Sum these squares and find the square root:
[tex]\[ \frac{5}{12} + \frac{7}{12} = 1 \][/tex]
[tex]\[ \sqrt{1} = 1 \][/tex]
- Since the magnitude is 1, this vector is a unit vector.
4. Fourth vector:
[tex]\[ u = \left\langle \frac{3}{\sqrt{7}}, -\frac{2}{\sqrt{7}} \right\rangle \][/tex]
- Calculate each component squared:
[tex]\[ \left(\frac{3}{\sqrt{7}}\right)^2 = \frac{9}{7} \][/tex]
[tex]\[ \left(-\frac{2}{\sqrt{7}}\right)^2 = \frac{4}{7} \][/tex]
- Sum these squares and find the square root:
[tex]\[ \frac{9}{7} + \frac{4}{7} = \frac{13}{7} \][/tex]
[tex]\[ \sqrt{\frac{13}{7}} \neq 1 \][/tex]
- Since the magnitude is not 1, this vector is not a unit vector.
In conclusion, the correct answers are:
[tex]\[ \boxed{2 \text{ and } 3} \][/tex]
