MHB Find Vectors $\vec{v_1}$, $\vec{v_2}$, $\vec{v_3}$ Given Dot Products

  • Thread starter Thread starter Saitama
  • Start date Start date
  • Tags Tags
    Vectors
AI Thread Summary
The discussion focuses on finding three vectors, $\vec{v_1}$, $\vec{v_2}$, and $\vec{v_3}$, based on given dot products. It is established that $\vec{v_1}$ can be chosen with a length of 2, while $\vec{v_2}$ must have a length of $\sqrt{2}$ and a specific dot product with $\vec{v_1}$. One participant suggests selecting $\vec{v_1} = (2,0,0)$ and finds that $\vec{v_2}$ can be determined as $(-1, 1, 0)$. The conversation concludes with the understanding that the vectors can be mirrored or rotated to generate additional solutions. The problem-solving approach emphasizes the flexibility in choosing the vectors while adhering to the constraints provided.
Saitama
Messages
4,244
Reaction score
93
Problem:
Find $\vec{v_1}$,$\vec{v_2}$ and $\vec{v_3}$ given that:

$$\vec{v_1}\cdot \vec{v_1}=4$$
$$\vec{v_1}\cdot \vec{v_2}=-2$$
$$\vec{v_1}\cdot \vec{v_3}=6$$
$$\vec{v_2}\cdot \vec{v_2}=2$$
$$\vec{v_2}\cdot \vec{v_3}=-5$$
$$\vec{v_3}\cdot \vec{v_3}=29$$

Attempt:
Assuming the vector $\vec{v_i}$ as $x_i \hat{i}+y_i \hat{j}+z_i \hat{k}$ is definitely not a good idea.

I am really clueless on how to tackle this problem. I can see that adding the second and fourth equation gives $\vec{v_2}\cdot (\vec{v_1}+\vec{v_2})=0$. This means that $\vec{v_1}$ is perpendicular to $\vec{v_1}+\vec{v_2}$ but I am not sure if this helps. I need a few hints to begin with.

Any help is appreciated. Thanks!
 
Mathematics news on Phys.org
Pranav said:
Problem:
Find $\vec{v_1}$,$\vec{v_2}$ and $\vec{v_3}$ given that:

$$\vec{v_1}\cdot \vec{v_1}=4$$
$$\vec{v_1}\cdot \vec{v_2}=-2$$
$$\vec{v_1}\cdot \vec{v_3}=6$$
$$\vec{v_2}\cdot \vec{v_2}=2$$
$$\vec{v_2}\cdot \vec{v_3}=-5$$
$$\vec{v_3}\cdot \vec{v_3}=29$$

Attempt:
Assuming the vector $\vec{v_i}$ as $x_i \hat{i}+y_i \hat{j}+z_i \hat{k}$ is definitely not a good idea.

I am really clueless on how to tackle this problem. I can see that adding the second and fourth equation gives $\vec{v_2}\cdot (\vec{v_1}+\vec{v_2})=0$. This means that $\vec{v_1}$ is perpendicular to $\vec{v_1}+\vec{v_2}$ but I am not sure if this helps. I need a few hints to begin with.

Any help is appreciated. Thanks!

Hey Pranav! ;)

There is a bit of freedom here.
You can pick any vector for $\vec v_1$ as long as its length is $2$.
After that, you can pick any vector for $\vec v_2$ as long as its length is $\sqrt 2$ and its dot product with $\vec v_1$ is $-2$.
After that we can see what remains for $\vec v_3$.

So let's pick $\vec v_1 = (2,0,0)$. That is with a zero component in both y- and z-directions.
And let's pick the z-coordinate of $\vec v_2$ zero.
Can you find a $\vec v_2$ that fits?
And after that a $\vec v_3$?

When you have found a solution, you can rotate or mirror that solution anyway you want (or a combination of those), and you'll have another solution.
 
I like Serena said:
Hey Pranav! ;)

There is a bit of freedom here.
You can pick any vector for $\vec v_1$ as long as its length is $2$.
After that, you can pick any vector for $\vec v_2$ as long as its length is $\sqrt 2$ and its dot product with $\vec v_1$ is $-2$.
After that we can see what remains for $\vec v_3$.

So let's pick $\vec v_1 = (2,0,0)$. That is with a zero component in both y- and z-directions.
And let's pick the z-coordinate of $\vec v_2$ zero.
Can you find a $\vec v_2$ that fits?
And after that a $\vec v_3$?
When you have found a solution, you can rotate or mirror that solution anyway you want (or a combination of those), and you'll have another solution.

Hi ILS! :)

Let $\vec{v_2}=x_2\hat{i}+y_2\hat{j}$. Then, from the second equation we have $2x_2=-2 \Rightarrow x_2=-1$, hence $y_2$ must be either $1$ or $-1$. Let us pick $y_2=1$.

I similarly found $\vec{v_3}$ (with $y_2=1$) equal to $3\hat{i}-2\hat{j}+4\hat{k}$.

...(or a combination of those), and you'll have another solution.
I am not sure what this means. Do you say that $v_1+v_2=\hat{i}+\hat{j}$ is also a solution to the given equation?

Thanks!
 
Pranav said:
Hi ILS! :)

Let $\vec{v_2}=x_2\hat{i}+y_2\hat{j}$. Then, from the second equation we have $2x_2=-2 \Rightarrow x_2=-1$, hence $y_2$ must be either $1$ or $-1$. Let us pick $y_2=1$.

I similarly found $\vec{v_3}$ (with $y_2=1$) equal to $3\hat{i}-2\hat{j}+4\hat{k}$.

Good!
I am not sure what this means. Do you say that $v_1+v_2=\hat{i}+\hat{j}$ is also a solution to the given equation?

Thanks!

No. I mean that you can mirror all three vectors, and then rotate all three vectors, and the result will be another solution.
If you rotate all 3 vectors together, their lengths and their directions relative to each other remain the same.
 
I like Serena said:
No. I mean that you can mirror all three vectors, and then rotate all three vectors, and the result will be another solution.
If you rotate all 3 vectors together, their lengths and their directions relative to each other remain the same.

Thanks a lot ILS! :)
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top