Linear Algebra: Finding a unit vector sharing a given angle with two other vectors

[haackeDc]

Find unit vector with a given angle to two other vectors in 3-space

Homework Statement

We are given the vectors <1,0,-1> and <0,1,1>, and are told to find a unit vector that shares an angle of (pi/3) with both of these vectors.

Homework Equations

a(dot)b = |a||b|cosθ

The Attempt at a Solution

So, from the information givin, the only thing I could think to do was form a system of linear equations:

u(dot)<1,0,-1> = (√2)cos(pi/3)
u(dot)<0,1,1> = (√2)cos(pi/3)

u1 - u2 = (.5)(√2)
u2 + u3 = (.5)(√2)

giving us:

u1 = -u3 + (√2)
u2 = -u3 + (.5)(√2)

so I end up with u = <-u3 + (√2), -u3 + (.5)(√2), u3>

u = u3<-1, -1, 1>


Now... this is as far away from my answer as I can be! I know it's not right, because this is the equation for a line, not a vector!

For informational purposes, the answer in the back of the book is <1/(√2), 1/(√2), 0>

How do I get to that answer?

 

[haackeDc]

Homework Statement

We are given the vectors <1,0,-1> and <0,1,1>, and are told to find a unit vector that shares an angle of (pi/3) with both of these vectors.

Homework Equations

a(dot)b = |a||b|cosθ

The Attempt at a Solution

So, from the information givin, the only thing I could think to do was form a system of linear equations:

u(dot)<1,0,-1> = (√2)cos(pi/3)
u(dot)<0,1,1> = (√2)cos(pi/3)

u1 - u2 = (.5)(√2)
u2 + u3 = (.5)(√2)

giving us:

u1 = -u3 + (√2)
u2 = -u3 + (.5)(√2)

so I end up with u = <-u3 + (√2), -u3 + (.5)(√2), u3>

u = (√2,√2, 0) + u3<-1, -1, 1>I know this isn't right, because this is the equation for a line, not a vector!

For informational purposes, the answer in the back of the book is <1/(√2), 1/(√2), 0>

How do I get to that answer?

 

[gneill]

Why not start by first reducing the two given vectors to unit vectors? That way the magnitudes become unity and drop out of the dot products. Let's call then u1 and u2. Also, since the pi/3 is 60°, the cosine is 1/2.

Suppose that u is the unknown unit vector. The dot products become

$u \cdot u1 = 1/2$
$u \cdot u2 = 1/2$

Let x,y, and z be the components of u. Then after expanding the dot products you should end up with:

$x - z = \frac{1}{\sqrt{2}}$
$y - z = \frac{1}{\sqrt{2}}$

Write the vector [x,y,z] parametrically. That is, suppose you write z = t. Then solve for x and y in terms of t. For what value of t will the resulting vector be a unit vector? What is that vector?

 

[lanedance]

note that if you find a u that satisfies you creteria, so will k u for any real k>0

so one other assumption you can make is that |u|^2=1

in fact you need it to get to the following
u(dot)<1,0,-1> = u1-u3 = |u|(√2)cos(pi/3) = (√2)/2 = 1/(√2)
u(dot)<0,1,1> = u2+u3 = |u|(√2)cos(pi/3) = (√2)/2 = 1/(√2)

rearranging
u1 = 1/(√2)+u3
u2 = 1/(√2)-u3

so then your vector is
<1/(√2)+u3, 1/(√2)-u3, u3>

now rather than considering as a line, why not use the last bit of info that |u|=1 to solve for u3

 

[SammyS]

Homework Statement

We are given the vectors <1,0,-1> and <0,1,1>, and are told to find a unit vector that shares an angle of (pi/3) with both of these vectors.

Homework Equations

a(dot)b = |a||b|cosθ

The Attempt at a Solution

So, from the information given, the only thing I could think to do was form a system of linear equations:

u(dot)<1,0,-1> = (√2)cos(pi/3)
u(dot)<0,1,1> = (√2)cos(pi/3)

u1 - u2 = (.5)(√2)     This should be u₁ - u₃ = (.5)(√2) .
u2 + u3 = (.5)(√2)

giving us:

u1 = -u3 + (√2)
u2 = -u3 + (.5)(√2)

so I end up with u = <-u3 + (√2), -u3 + (.5)(√2), u3>

u = (√2,√2, 0) + u3<-1, -1, 1>


I know this isn't right, because this is the equation for a line, not a vector!

For informational purposes, the answer in the back of the book is <1/(√2), 1/(√2), 0>

How do I get to that answer?

You have a typo or simply a mistake.

You can think of your result as the equation of a line, but it's the vector equation if the line. That's essentially a family of position vectors, which terminate at various locations along the line. You want the one or two of these position vectors which have a magnitude of 1 , as lancedance pointed out.

 

[haackeDc]

note that if you find a u that satisfies you creteria, so will k u for any real k>0

so one other assumption you can make is that |u|^2=1

in fact you need it to get to the following
u(dot)<1,0,-1> = u1-u3 = |u|(√2)cos(pi/3) = (√2)/2 = 1/(√2)
u(dot)<0,1,1> = u2+u3 = |u|(√2)cos(pi/3) = (√2)/2 = 1/(√2)

rearranging
u1 = 1/(√2)+u3
u2 = 1/(√2)-u3

so then your vector is
<1/(√2)+u3, 1/(√2)-u3, u3>

now rather than considering as a line, why not use the last bit of info that |u|=1 to solve for u3

lanedance, thank you.

So... interesting thought: if you were only given one reference vector, would the solution set be the shell of a cone?

EDIT: If you weren't given the restriction that it had to be a unit vector

 

[SammyS]

I'm not sure if you noticed, but the angle between <1,0,-1> and <0,1,1> is 2π/3 .

This means that there is only one solution to your problem, and all three vectors lie in the same plane.

If the angle between <1,0,-1> and <0,1,1> were less than 2π/3, there would be two solutions.

 

[lanedance]

if you remove the magnitude restraint the solution set would be two separate half lines, as any positive multiple of a vector will form the same angle

 

[lanedance]

SammyS doesn't that only add to 4pi/3 in the plane?

 

[haackeDc]

if you remove the magnitude restraint the solution set would be two separate half lines, as any positive multiple of a vector will form the same angle

I'm talking about if you were only given a SINGLE REFERENCE VECTOR, not two, and then told to describe the solution set of all vectors from a given angle to that single vector. It would be a cone, wouldn't it?

 

[lanedance]

correct

 

[SammyS]

I'm not sure if you noticed, but the angle between <1,0,-1> and <0,1,1> is 2π/3 .

This means that there is only one solution to your problem, and all three vectors lie in the same plane.

If the angle between <1,0,-1> and <0,1,1> were less than 2π/3, there would be two solutions.

#@!&%$ TYPO !

I even thought to myself ... Check for a typo before posting this ! DUH !

The angle between <1,0,-1> and <0,1,1> is 4π/3 ... !
...

If the angle between the two given vectors was less than 4π/3, there would be two solutions.