# Finding the vectors

#### Pranav

##### Well-known member
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!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.

#### Pranav

##### Well-known member
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!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.

#### Pranav

##### Well-known member
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!