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

In summary, the problem involves three non-coplanar unit vectors that are equally inclined to each other at an angle $\theta$, and the goal is to find the scalars $p$, $q$, and $r$ in terms of $\theta$. The attempt involved taking the dot product of both sides of the equation successively with each vector, resulting in three equations. These equations were then used to find the volume of the parallelepiped spanned by the three vectors, and it was determined that the scalar triple product is equal to $(1-\cos\theta)\sqrt{2\cos\theta+1}$. To solve for $p$, $q$, and $r$, Cramer's rule or an alternative method involving taking the
  • #1
Saitama
4,243
93
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!
 
Last edited:
Mathematics news on Phys.org
  • #2
Pranav said:
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 $\theta$?
 
  • #3
Hello again! :)

I like Serena said:
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 $\vec{b}\times \vec{c}$. :confused:
 
  • #4
Pranav said:
But to find the volume I need the angle between $\vec{a}$ and $\vec{b}\times \vec{c}$. :confused:

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...
 
  • #5
I like Serena said:
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?
 
  • #6
Pranav said:
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 $\sin\theta\sin\alpha$...
 
  • #7
I like Serena said:
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. :eek:

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? :confused:
 
  • #8
Pranav said:
Yes, you are right. :eek:

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? :confused:

Cramer's rule seems like a good idea.
I don't see a shortcut.
 
  • #9
Pranav said:
...

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? :confused:
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.
 
  • #10
Opalg said:
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. :confused:
 
  • #11
Pranav said:
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.
 
  • #12
Opalg said:
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. :confused:
 
  • #13
Pranav said:
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. :confused:

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.
 
  • #14
I like Serena said:
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
$$p=-\frac{1}{\sqrt{2\cos\theta+1}}, q=\frac{2q}{\sqrt{2\cos\theta+1}}, r=-\frac{1}{\sqrt{2\cos\theta+1}}$$
 

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

1. What is the purpose of using vector algebra in problem #3?

Vector algebra is used to represent and analyze mathematical quantities that have both magnitude and direction in problem #3. It allows for the manipulation and calculation of these quantities, making it a powerful tool in solving problems involving vectors.

2. What are the basic operations in vector algebra?

The basic operations in vector algebra include addition, subtraction, scalar multiplication, and dot product. These operations allow for the combining and comparison of vectors to solve problems.

3. How do you determine the magnitude of a vector in problem #3?

The magnitude of a vector can be determined using the Pythagorean theorem, which states that the magnitude of a vector is equal to the square root of the sum of the squares of its components. In problem #3, the magnitude of a vector can be calculated using this formula to solve for its length.

4. What is the difference between a scalar and a vector in problem #3?

A scalar is a mathematical quantity that has only magnitude, while a vector has both magnitude and direction. In problem #3, scalars can be represented by numerical values, while vectors are represented by a magnitude and direction.

5. How do you determine the direction of a vector in problem #3?

The direction of a vector can be determined using trigonometric functions, such as sine and cosine. In problem #3, the direction of a vector can be found by calculating the angle between the vector and a given reference direction, usually the positive x-axis.

Back
Top