# Find two unit vectors that make an angle of 45° with the vector v = (4, 3).

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:

$v_1 = \sqrt{ v1_x^2 + v1_y^2 }$

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 the 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:

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

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

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

$19w_x^2\ -\ 28.28w_x + 9.5 = 0$

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:

$< 0.8, -0.6 > , < -0.8, 0.6 >$