Welcome to our community

Be a part of something great, join today!

Problem of the Week #1 - April 2nd, 2012

Status
Not open for further replies.
  • Thread starter
  • Admin
  • #1

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Thanks to Chris L T521 for submitting this problem!

Let $\mathbf{u}=u_1\mathbf{i}+u_2\mathbf{j}+u_3\mathbf{k}$ and $\mathbf{v}=v_1\mathbf{i}+v_2\mathbf{j}+v_3\mathbf{k}$ be two vectors in $\mathbb{R}^3$. We define the norm of the vector $\mathbf{v}$ by
\[\|\mathbf{v}\| = \sqrt{v_1^2+v_2^2+v_3^2},\]
the dot product of $\mathbf{u}$ and $\mathbf{v}$ by
\[\mathbf{u}\cdot\mathbf{v}=u_1v_1 + u_2v_2 + u_3v_3,\]
and the cross product of $\mathbf{u}$ and $\mathbf{v}$ by
\[\mathbf{u}\times\mathbf{v} = \det\begin{pmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3\\ v_1 & v_2 & v_3\end{pmatrix}.\]

Express $\|\mathbf{u}\times\mathbf{v}\|^2 + (\mathbf{u}\cdot \mathbf{v})^2$ in terms of $\|\mathbf{u}\|$ and $\|\mathbf{v}\|$ only.

No hints for this one either! ;)

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
Last edited:
  • Thread starter
  • Admin
  • #2

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Congratulations to the following members for their correct solutions:

1) grgrsanjay
2) Sudharaka


Solution below:

We observe that $\mathbf{u}\times\mathbf{v} = (u_2v_3 - v_2u_3)\mathbf{i} - (u_1v_3 - v_1u_3)\mathbf{j} + (u_1v_2-v_1u_2)\mathbf{k}$.

Thus,

\[\begin{aligned}\|\mathbf{u}\times\mathbf{v}\|^2 &= (u_2v_3-v_2u_3)^2 + (u_1v_2-v_1u_3)^2 + (u_1v_2-v_1u_2)^2\\ &= (u_2^2v_3^2-2u_2u_3v_2v_3 + v_2^2u_3^2) +(u_1^2v_3^2 - 2u_1u_3v_1v_3 + v_1^2u_3^2)+(u_1^2v_2^2-2u_1u_2v_1v_2+v_1^2u_2^2)\end{aligned}\]

Since $\mathbf{u}\cdot\mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3$, we see that

\[\begin{aligned}(\mathbf{u}\cdot\mathbf{v})^2 &= u_1v_1(u_1v_1 + u_2v_2+u_3v_3) + u_2v_2(u_1v_1 + u_2v_2+u_3v_3) + u_3v_3(u_1v_1 + u_2v_2+u_3v_3)\\ &= u_1^2v_1^2 + u_2^2v_2^2 +u_3v_3^2+2u_1u_2v_1v_2 + 2u_1u_3v_1v_3 + 2u_2u_3v_2v_3\end{aligned}\]

So, it now follows that

\[\begin{aligned}\|\mathbf{u}\times\mathbf{v}\|^2+( \mathbf{u}\cdot\mathbf{v})^2 &= u_1^2v_1^2+u_1^2v_2^2 + u_1^2v_3^2 + u_2^2v_1^2 + u_2^2v_2^2 + u_2^2v_3^2 + u_3^2v_1^2+u_3^2v_2^2 + u_3^2v_3^2\\ &= u_1^2(v_1^2+v_2^2+v_3^2) + u_2^2(v_1^2+v_2^2+v_3^2)+u_3^2(v_1^2+v_2^2+v_3^2) \\ &= (u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2). \end{aligned}\]

Thus, $\|\mathbf{u}\times\mathbf{v}\|^2+(\mathbf{u}\cdot\mathbf{v})^2 = \|\mathbf{u}\|^2\|\mathbf{v}\|^2$.

If you submitted a solution and your name isn't listed, please check your PM box. We have most likely messaged you asking for some clarification.
 
Last edited:
Status
Not open for further replies.