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.

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

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

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.