The question aims to find **two unit vectors** that make an **angle** of $45^{\circ}$ with the given **vector v**.The question depends on the concept of **unit vectors,** the **dot product** between two vectors, and the **length** of a **vector.** The **length** of the **vector** is also its **magnitude.** The length of a** 2D vector** is given as:

\[ v_1 = \sqrt{ v1_x^2 + v1_y^2 } \]

## Expert Answer

The given vector is:

\[ v = (4, 3) \]

We need to find **two unit vectors** that make an angle of $45^{\circ}$ with the given vector. To find those **vectors,** we need to take theÂ **dot product** of the vector with an unknown **vector** and use the yielded equation to find the vectors.

Let us assume the **unit vector** is **w** and its **magnitude** is given as:

\[ |w| = \sqrt{ w_x^2 + w_y^2 } \]

\[ |w| = 1 \]

The **dot product** of the vectors is given as:

\[ v . w = \sqrt{ 4^2 + 3^2 } . 1 \cos \theta \]

\[ < 4, 3 > . < w_x, w_y > = \sqrt{25} \cos (45) \]

\[ 4w_x + 3w_y = (5) \dfrac {1} {\sqrt{2}} \]

\[ 4w_x + 3w_y = 3.535 \]

\[ w_y = \dfrac{ 3.535\ -\ 4w_x }{ 3 } \hspace{1in} (1) \]

As the **magnitude** of the **unit vector** is given as:

\[ \sqrt{ w_x^2 + w_y^2 } = 1 \]

\[ w_x^2 + w_y^2 = 1 \]

Substituting the value of $w_y$ in the above equation, we get:

\[ w_x^2 + (Â \dfrac{ 3.535\ -\ 4w_x }{ 3 } )^2 = 1 \]

\[ 3w_x^2 + (3.535\ -\ w_x)^2 -\ 3 = 0 \]

\[ 3w_x^2 + 12.5 + 16w_x^2\ -\ 2 (3.535) (4w_x)\ -\ 3 = 0 \]

\[ 19w_x^2\ -\ 28.28w_x + 9.5 = 0 \]

Using the **quadratic equation,** we get:

\[ w_x = [ 0.98, 0.51 ] \]

Using these values of **$’w_x’$** in equation (1), we get:

\[ w_y = \dfrac{ 3.535\ -\ 4(0.98) }{ 3 } \]

\[ w_y = – 0.1283 \]

The **first unit vector** is calculated to be:

\[ < 0.98, -0.1283 > \]

\[ w_y = \dfrac{ 3.535\ -\ 4(0.51) }{ 3 } \]

\[ w_y = 0.4983 \]

The **second unit vector** is calculated to be:

\[ < 0.51, 0.4983 > \]

## Numerical Result

The **first unit vector** is calculated to be:

\[ < 0.98, -0.1283 > \]

The **second unit vector** is calculated to be:

\[ < 0.51, 0.4983 > \]

## Example

Find a **unit vectors perpendicular** to the **vector v = <3, 4>.**

The **magnitude** of the **unit vector** is given as:

\[ |u| = \sqrt{ x^2 + y^2 } \]

\[ |u| = 1 \]

\[ x^2 + y^2 = 1 \]

The **dot product** of the **vectors perpendicular** to each other is given as:

\[ u . v = |u| |v| \cos (90) \]

\[ u . v = 0 \]

\[ < 3, 4 > . < x, y > = 0 \]

\[ 3x + 4y = 0 \]

\[ y = – \dfrac {3} {4} x \]

Substituting the value of **y** in the above equation, we get:

\[ x^2 + (- \dfrac {3} {4} x )^2 = 1 \]

\[ x^2 + \dfrac{9}{16} x^2 = 1 \]

\[ 1.5625x^2 = 1 \]

\[ x^2 = \dfrac{ 1 }{ 1.5625 } \]

\[ x^2 = 0.64 \]

\[ x = \pm \sqrt{0.64} \]

\[ x = \pm 0.8 \]

The vectors **perpendicular** to the given **vectors** are:

\[ < 0.8, -0.6 > , < -0.8, 0.6 > \]