# Find a unit vector for a particular vector

1. Jan 13, 2016

### RyanTAsher

1. The problem statement, all variables and given/known data

Find two unit vectors in 2-space that make an angle of 45(deg) with 7i + 6j.

2. Relevant equations

Unit Vector = u = (1/||v||)*v

Dot Product = u . v = ||u|| ||v|| cos(theta)

3. The attempt at a solution

I've been thinking about a way to do this. I originally thought that unit vectors could only be in the direction of the vector specified, because direction was the only thing that mattered.

I was thinking, if I found the unit vector of 7i + 6j, and then I multiplied that by cos(45), and sin(45), it would give me two solutions of unit vectors 45 degrees from the direction of 7i + 6j.

I'm not sure if this thinking is correct.

2. Jan 13, 2016

### LCKurtz

Hint: First find two vectors perpendicular to 7i+6j. What direction do you get if you add two equal length perpendicular vectors?

3. Jan 14, 2016

### RyanTAsher

Ahhh, okay I was thinking you would find the unit vectors first for some reason.

So, we are just finding 2 vectors directed at 45 degrees from the original vector, and then finding the unit vectors along those directions.

Thank you.

4. Jan 15, 2016

### RyanTAsher

Sorry to bring this question back up, but I just wanted to confirm my answer, as my math question online program I think might be reading my answer wrong.

$\vec{v} = 7 \hat{i} + 6\hat{j}$

Let a vector 45 degrees from $\vec{v}$ be equal to $\vec{w} = a\hat{i} + b\hat{j}$.

$cos(\theta) = \frac {\vec{v} \bullet \vec{w}} {||v|| ||w||}$

$\frac {\sqrt{2}} {2} = \frac {\langle 7, 6 \rangle \bullet \langle a, b \rangle} {\sqrt{85} \sqrt{a^{2} + b^{2}}}$

$\frac {\sqrt{2}} {2} = \frac {7a + 6b} {\sqrt{85} \sqrt{a^{2} + b^{2}}}$

$\sqrt{2} \sqrt{85} \sqrt{a^{2} + b^{2}} = 2(7a + 6b)$

$(\sqrt{2} \sqrt{85} \sqrt{a^{2} + b^{2}})^{2} = (2(7a + 6b))^{2}$

$(170)(a^{2} + b^{2}) = 4(7a + 6b)^{2}$

Let a = 1.

$(170)(1 + b^{2}) = 4(7 + 6b)^{2}$
$170 + 170b^{2} = 4(49 + 84b + 36b^{2})$
$170 + 170b^{2} = 196 + 336b + 144b^{2}$
$26b^{2} - 336b - 26 = 0$

Quadratic equation gives:

$b = \frac {84 ± 85} {13}$
$b = 13, b = - \frac {1} {13}$

Therefore, the two vectors that are 45 degrees from v, are:

$w_1 = \langle 1, 13 \rangle$
$w_2 = \langle 1, -\frac {1} {13} \rangle$

The unit vectors are then:

$u_1 = \frac {\langle 1, 13 \rangle} {\sqrt{170}}$
$u_2 = \frac {\langle 1, -\frac {1} {13} \rangle} {\sqrt{1 + \frac {1} {169}}}$

5. Jan 15, 2016

### LCKurtz

Your answers are correct but I would expect your online software would expect you to simplify them, especially $u_2$. You have certainly gone the long way around Robin Hood's barn to get it. Given the vector $\langle 7,6\rangle$, isn't it obvious using the dot product that $\langle -6,7\rangle$ and its negative are perpendicular to it? Then you could have used my original hint. Still, you do have a solution and your answers are correct.

6. Jan 16, 2016

### ehild

@LCKurtz ingenious suggestion was to find a vector that is of equal length and perpendicular to the original one, and then add the two vectors. The resultant vector halves the angle between $\vec a$ and $\vec b$. See figure.
You get the blue vector if rotating $\vec a = 7\vec i + 6 \vec j$ by 90°. $\vec b = -6\vec i + 7 \vec j$. Adding them: $\vec a + \vec b = 1\vec i + 13 \vec j$. Now you have to divide it by the norm. Do the same with the other perpendicular vector, negative of $\vec b$.

7. Jan 16, 2016

### RyanTAsher

Ohhhh, thank you for that picture. I understand now. I knew that the 45 degree vector would be half the 90 degree, but I wasn't too sure on how to get it there. Thank you for the picture.

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