MHB How Do Scalars Relate to Unit Vectors Inclined at Angle Theta?

[Saitama]

Problem:
Let $\vec{a}$,$\vec{b}$ and $\vec{c}$ be non-coplanar unit vectors, equally incline to one another at an angle $\theta$. If $\vec{a}\times \vec{b} + \vec{b}\times \vec{c}=p\vec{a}+q\vec{b}+r\vec{c}$. Find scalars $p$,$q$ and $r$ in terms of $\theta$.

Attempt:
Taking the dot product on both sides successively with $\vec{a}$,$\vec{b}$ and $\vec{c}$, I get the following three equations:
$$\begin{array}
\\
[\vec{a} \vec{b} \vec{c}]=p+q\cos\theta+r\cos\theta \\
0=p\cos\theta+q+r\cos\theta \\
[\vec{a} \vec{b} \vec{c}]=p\cos\theta+q\cos\theta+r \\
\end{array}
$$

I am not sure how to proceed after this. I guess I have to express $[\vec{a} \vec{b} \vec{c}]$ in terms of $\theta$ but I don't see how.

Any help is appreciated. Thanks!

 

[I like Serena]

Problem:
Let $\vec{a}$,$\vec{b}$ and $\vec{c}$ be non-coplanar unit vectors, equally incline to one another at an angle $\theta$. If $\vec{a}\times \vec{b} + \vec{b}\times \vec{c}=p\vec{a}+q\vec{b}+r\vec{c}$. Find scalars $p$,$q$ and $r$ in terms of $\theta$.

Attempt:
Taking the dot product on both sides successively with $\vec{a}$,$\vec{b}$ and $\vec{c}$, I get the following three equations:
$$\begin{array}
\\
[\vec{a} \vec{b} \vec{c}]=p+q\cos\theta+r\cos\theta \\
0=p\cos\theta+q+r\cos\theta \\
[\vec{a} \vec{b} \vec{c}]=p\cos\theta+q\cos\theta+r \\
\end{array}
$$

I am not sure how to proceed after this. I guess I have to express $[\vec{a} \vec{b} \vec{c}]$ in terms of $\theta$ but I don't see how.

Any help is appreciated. Thanks!

Hey Pranav! :)

The expression $[\vec{a} \vec{b} \vec{c}]$ is equal to the volume of the parallelepiped that is the span of the 3 vectors.
Can you also find this volume in terms of $\theta$?

 

[Saitama]

Hello again! :)

The expression $[\vec{a} \vec{b} \vec{c}]$ is equal to the volume of the parallelepiped that is the span of the 3 vectors.
Can you also find this volume in terms of $\theta$?

But to find the volume I need the angle between $\vec{a}$ and $\vec{b}\times \vec{c}$.

 

[I like Serena]

But to find the volume I need the angle between $\vec{a}$ and $\vec{b}\times \vec{c}$.

Yes.
That is the angle between $\vec{a}$ and the triangle between $\vec{b}$ and $\vec{c}$.
That last triangle is an isosceles triangle.
If you could find a vector that divides its angle in 2, you might take the angle of $\vec{a}$ with that vector...

 

[Saitama]

Yes.
That is the angle between $\vec{a}$ and the triangle between $\vec{b}$ and $\vec{c}$.
That last triangle is an isosceles triangle.
If you could find a vector that divides its angle in 2, you might take the angle of $\vec{a}$ with that vector...

The angle bisector vector of $\vec{b}$ and $\vec{c}$ is given by $\vec{b}+\vec{c}$. The angle ($\alpha$) between this vector and $\vec{a}$ is given by:
$$\cos\alpha=\frac{\vec{a}\cdot (\vec{b}+\vec{c})}{|\vec{b}+\vec{c}|}$$
$$\Rightarrow \cos\alpha=\frac{2\cos\theta}{\sqrt{2+2\cos\theta}}=\frac{\sqrt{2}\cos\theta}{\sqrt{1+\cos\theta}}$$

The volume is then given by:
$$[\vec{a} \vec{b} \vec{c}]=\cos\theta \sin(\pi/2-\alpha)=\frac{\sqrt{2}\cos^2\theta}{\sqrt{1+\cos \theta}}$$

Is this correct?

 

[I like Serena]

The angle bisector vector of $\vec{b}$ and $\vec{c}$ is given by $\vec{b}+\vec{c}$. The angle ($\alpha$) between this vector and $\vec{a}$ is given by:
$$\cos\alpha=\frac{\vec{a}\cdot (\vec{b}+\vec{c})}{|\vec{b}+\vec{c}|}$$
$$\Rightarrow \cos\alpha=\frac{2\cos\theta}{\sqrt{2+2\cos\theta}}=\frac{\sqrt{2}\cos\theta}{\sqrt{1+\cos\theta}}$$

Good!

The volume is then given by:
$$[\vec{a} \vec{b} \vec{c}]=\cos\theta \sin(\pi/2-\alpha)=\frac{\sqrt{2}\cos^2\theta}{\sqrt{1+\cos \theta}}$$

That does not look right.

The area of parallelogram b,c is $bc\sin\theta = \sin\theta$.
The height of the parallelepiped is $a\sin\alpha = \sin\alpha$.
That makes the volume of the parallelepiped $\sin\theta\sin\alpha$...

 

[Saitama]

That does not look right.

The area of parallelogram b,c is $bc\sin\theta = \sin\theta$.
The height of the parallelepiped is $a\sin\alpha = \sin\alpha$.
That makes the volume of the parallelepiped $\sin\theta\sin\alpha$...

Yes, you are right.

I should have written
$$[\vec{a} \vec{b} \vec{c}]=\sin\theta \cos(\pi/2-\alpha)=\sin\theta \sin\alpha$$
Since I found $\cos\alpha$ in my previous post,
$$\Rightarrow \sin\alpha =\sqrt{1-\frac{2\cos^2\theta}{1+\cos\theta}}=\sqrt{\frac{(2\cos\theta+1)(1-\cos\theta)}{(1+\cos\theta)}}=\frac{1-\cos\theta}{\sin\theta}\sqrt{2\cos\theta+1}$$
Hence, the scalar triple product is
$$[\vec{a} \vec{b} \vec{c}]=(1-\cos\theta)\sqrt{2\cos\theta+1}$$
Next comes solving the linear equations. Should I go by Cramer's rule or is their an easier shortcut?

 

[I like Serena]

Yes, you are right.

I should have written
$$[\vec{a} \vec{b} \vec{c}]=\sin\theta \cos(\pi/2-\alpha)=\sin\theta \sin\alpha$$
Since I found $\cos\alpha$ in my previous post,
$$\Rightarrow \sin\alpha =\sqrt{1-\frac{2\cos^2\theta}{1+\cos\theta}}=\sqrt{\frac{(2\cos\theta+1)(1-\cos\theta)}{(1+\cos\theta)}}=\frac{1-\cos\theta}{\sin\theta}\sqrt{2\cos\theta+1}$$
Hence, the scalar triple product is
$$[\vec{a} \vec{b} \vec{c}]=(1-\cos\theta)\sqrt{2\cos\theta+1}$$

I get the same! :D

Next comes solving the linear equations. Should I go by Cramer's rule or is their an easier shortcut?

Cramer's rule seems like a good idea.
I don't see a shortcut.

 

[Opalg]

...

Hence, the scalar triple product is
$$[\vec{a} \vec{b} \vec{c}]=(1-\cos\theta)\sqrt{2\cos\theta+1}$$
Next comes solving the linear equations. Should I go by Cramer's rule or is their an easier shortcut?

If you take the vector product of both sides of the original equation with $\vec b$ then you get $0 = (p-r)\sin\theta$, from which it follows that $p=r$. Then your three equations reduce to $[\vec{a} \vec{b} \vec{c}]=p(1+\cos\theta)+q\cos\theta$ and $q = -2p\cos\theta,$ together with $r=p$. Those are easy to solve.

 

[Saitama]

If you take the vector product of both sides of the original equation with $\vec b$ then you get $0 = (p-r)\sin\theta$, from which it follows that $p=r$. Then your three equations reduce to $[\vec{a} \vec{b} \vec{c}]=p(1+\cos\theta)+q\cos\theta$ and $q = -2p\cos\theta,$ together with $r=p$. Those are easy to solve.

That looks a lot better than Cramer's rule but I don't seem to be getting the same equation as you. :(

Taking the vector product with $\vec{b}$ on LHS gives:
$$\vec{b}\times (\vec{a} \times \vec{b})+\vec{b}\times (\vec{b}\times \vec{c})=\vec{a}(\vec{b}\cdot \vec{b})-\vec{b}(\vec{a}\cdot \vec{b})+\vec{b}(\vec{b}\cdot \vec{c})-\vec{c}(\vec{b}\cdot \vec{b})=\vec{a}-\vec{c}$$

But you get zero on the left side.

 

[Opalg]

That looks a lot better than Cramer's rule but I don't seem to be getting the same equation as you. :(

Sorry, ignore my previous comment, which was nonsense. The reason that $p=r$ is simply that if you subtract the last of the three equations $$\begin{array}
\\
[\vec{a} \vec{b} \vec{c}]=p+q\cos\theta+r\cos\theta \\
0=p\cos\theta+q+r\cos\theta \\
[\vec{a} \vec{b} \vec{c}]=p\cos\theta+q\cos\theta+r \\
\end{array}
$$ from the first, then you get $0 = p(1-\cos\theta) - r(1 - \cos\theta)$. The equations then become easy to solve.

 

[Saitama]

Sorry, ignore my previous comment, which was nonsense. The reason that $p=r$ is simply that if you subtract the last of the three equations $$\begin{array}
\\
[\vec{a} \vec{b} \vec{c}]=p+q\cos\theta+r\cos\theta \\
0=p\cos\theta+q+r\cos\theta \\
[\vec{a} \vec{b} \vec{c}]=p\cos\theta+q\cos\theta+r \\
\end{array}
$$ from the first, then you get $0 = p(1-\cos\theta) - r(1 - \cos\theta)$. The equations then become easy to solve.

Thanks Opalg, I seem to get:
$$p=\frac{1}{\sqrt{2\cos\theta+1}}$$
But the answer key mentions one more solution, which is
$$p=-\frac{1}{\sqrt{2\cos\theta+1}}$$
I couldn't get this solution by solving the above equations.

 

[I like Serena]

Thanks Opalg, I seem to get:
$$p=\frac{1}{\sqrt{2\cos\theta+1}}$$
But the answer key mentions one more solution, which is
$$p=-\frac{1}{\sqrt{2\cos\theta+1}}$$
I couldn't get this solution by solving the above equations.

The volume of the parallelepiped is actually $\Big| [\vec a \vec b \vec c] \Big|$.
In other words, there is another solution with $\vec a, \vec b, \vec c$ ordered left-handed instead of right-handed.

 

[Saitama]

The volume of the parallelepiped is actual $\Big| [\vec a \vec b \vec c] \Big|$.
In other words, there is another solution if $\vec a, \vec b, \vec c$ are ordered left-handed instead of right-handed.

Yes, you are right. Thanks a lot ILS and Opalg, the problem has been solved. :)

The final answers:
$$p=\frac{1}{\sqrt{2\cos\theta+1}}, q=-\frac{2q}{\sqrt{2\cos\theta+1}}, r=\frac{1}{\sqrt{2\cos\theta+1}}$$
or
$$p=-\frac{1}{\sqrt{2\cos\theta+1}}, q=\frac{2q}{\sqrt{2\cos\theta+1}}, r=-\frac{1}{\sqrt{2\cos\theta+1}}$$