Finding a unit vector with the given properties

[Illania]

Homework Statement

The problem states to find a unit vector that is orthogonal to $\left\langle1, 1, -2\right\rangle$, forms an angle of $\frac{\pi}{4}$ with $\left\langle1, 1, 1\right\rangle$ and has v₁ > 0.

Homework Equations

$cos\theta = \frac{\vec{u}\bullet\vec{v}}{|\vec{u}||\vec{v}|}$

The Attempt at a Solution

Since $\vec{v}$ is a unit vector, I know the length is one. This means that in the above equation, the denominator will simply be $|\vec{u}|$.

I know that $\left\langle1, 1, -2\right\rangle \bullet$ $\left\langle v_{1}, v_{2}, v_{3}\right\rangle = 0$ so $v_{3} = \frac{v_{1}+v_{2}}{2}$

$cos(\frac{\pi}{4}) = \frac{\left\langle1, 1, 1\right\rangle \bullet \left\langle v_{1}, v_{2}, v_{3}\right\rangle}{\sqrt{3}}$ so

$v_{3} = \sqrt{\frac{3}{2}} - v_{1} - v_{2}$

I set the two equations for $v_{3}$ equal to each other and end up with $v_{1} + v_{2} = \frac{\sqrt{6}}{3}$ but I'm not sure where to go from here.

 

[Mark44]

Homework Statement

The problem states to find a unit vector that is orthogonal to $\left\langle1, 1, -2\right\rangle$, forms an angle of $\frac{\pi}{4}$ with $\left\langle1, 1, 1\right\rangle$ and has v₁ > 0.

Homework Equations

$cos\theta = \frac{\vec{u}\bullet\vec{v}}{|\vec{u}||\vec{v}|}$

The Attempt at a Solution

Since $\vec{v}$ is a unit vector, I know the length is one. This means that in the above equation, the denominator will simply be $|\vec{u}|$.

I know that $\left\langle1, 1, -2\right\rangle \bullet$ $\left\langle v_{1}, v_{2}, v_{3}\right\rangle = 0$ so $v_{3} = \frac{v_{1}+v_{2}}{2}$

$cos(\frac{\pi}{4}) = \frac{\left\langle1, 1, 1\right\rangle \bullet \left\langle v_{1}, v_{2}, v_{3}\right\rangle}{\sqrt{3}}$ so

$v_{3} = \sqrt{\frac{3}{2}} - v_{1} - v_{2}$

I set the two equations for $v_{3}$ equal to each other and end up with $v_{1} + v_{2} = \frac{\sqrt{6}}{3}$ but I'm not sure where to go from here.

I think you're missing one equation. v is a unit vector, so v₁² + v₂² + v₃² = 1.

 

[Illania]

I think you're missing one equation. v is a unit vector, so v₁² + v₂² + v₃² = 1.

Yes, you're right. I had forgotten about that. Even using that, though, I am still unsure of the next steps I should take. Is the previous work I showed applicable? Was I going in the wrong direction?

 

[haruspex]

I am still unsure of the next steps I should take.

You now know v₃, so you can calculate v₃², then v₁²+v₂², etc.

 

[Mark44]

You're heading in the right direction. With three equations and three unknowns, you should be able to solve for the coefficients of v.

You might have a mistake in your work in your second equation.

cos($\pi/4$) = √2 /2. I don't see how you got what you show for v₃ from your equation.

 

[haruspex]

You're heading in the right direction. With three equations and three unknowns, you should be able to solve for the coefficients of v.

You might have a mistake in your work in your second equation.

cos($\pi/4$) = √2 /2. I don't see how you got what you show for v₃ from your equation.

No, I think it's right. More obviously if you write cos($\pi/4$) = 1/√2

 

[Illania]

I think I may be missing something obvious here as I can't seem to figure out how to find v₃ from the work I have shown here.

 

[Mark44]

No, I think it's right. More obviously if you write cos($\pi/4$) = 1/√2

No, it's not right.

You have
$$\frac{v_1 + v_2 + v_3}{\sqrt{3}} = \frac{1}{\sqrt{2}} $$
$$ \Rightarrow v_1 + v_2 + v_3 = \frac{\sqrt{3}}{\sqrt{2}}$$

That's the same as √6/2, not √6/3 as you show.

Your first two equations are
$v_1 + v_2 - 2v_3 = 0$
$v_1 + v_2 + v_3 = \frac{\sqrt{3}}{\sqrt{2}}$

Subtract one of these from the other to solve for v₃.

Then substitute that value in the first equation above to solve for v₁ in terms of v₂.

Then substitute for v₁ and v₃ in the third equation.

 

[haruspex]

No, it's not right.

You have
$$\frac{v_1 + v_2 + v_3}{\sqrt{3}} = \frac{1}{\sqrt{2}} $$
$$ \Rightarrow v_1 + v_2 + v_3 = \frac{\sqrt{3}}{\sqrt{2}}$$

That's the same as √6/2, not √6/3 as you show.

√6/3 is the value obtained for v₁+v₂.

 

[Mark44]

√6/3 is the value obtained for v₁+v₂.

OK, I didn't catch that.

 

[haruspex]

$v_{3} = \frac{v_{1}+v_{2}}{2}$
$v_{1} + v_{2} = \frac{\sqrt{6}}{3}$
I can't seem to figure out how to find v₃ from the work I have shown here.

You can't see how to combine those two equations to find v₃?

 

[Illania]

You can't see how to combine those two equations to find v₃?

Ok, so I've found v₃ to be $\frac{\sqrt{6}}{6}$.

I plugged this back into $v_{1} + v_{2} -2v_{3} = 0$ to get $v_{2} = \frac{\sqrt{6}}{3} - v_{1}$.

I plugged this into $v_{1}^{2} + v_{2}^{2} + v_{3}^2 = 1$ and ended up with $2v_{1}^{2} - \frac{2\sqrt{6}}{3}v_{1} -\frac{1}{6} = 0$.

From this, I found that $v_{1} = \frac{\sqrt{6}}{6} \pm \frac{1}{2}$. I choose that $v_{1} = \frac{\sqrt{6}}{6} + \frac{1}{2}$ since it is greater than 0. From here I get that v₂ equals the same thing as v₁ using what I have above. However, when I use these values for $\vec{v}$, I get that $\vec{v}\bullet\left\langle1, 1, -2\right\rangle = 1$ rather than the 0 that it should be. I can't seem to catch my mistake here.

EDIT: I've found the mistake. It was a subtraction error. Thank you for the help, everyone.