# Find the Vector in the New Position Obtained by Rotation

The vector $\vec{OP}=\hat{i}+2\hat{j}+2\hat{k}$ turns through a right angle,passing through the positive $x-$axis on the way.Find the vector in the new position.

Let the new position of the vector be $OP'=x\hat{i}+y\hat{j}+z\hat{k}$.As $OP$ is making $90$ degrees with $OP'$.
So $(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}+2\hat{j}+2\hat{k})=0\Rightarrow x+2y+2z=0$
Also $|\vec{OP}|=|\vec{OP'}|$
$x^2+y^2+z^2=9$
Now i have only two equations and three variables to find.I cannot solve to find $x,y,z$.I dont to how to use the statement "passing through the positive $x-$axis on the way".Maybe that can give me the required equation.

• "turns through a right angle" about which axis??? or which vector???? Info required. – SchrodingersCat Nov 5 '15 at 12:38
• What is the axis of rotation? :) – H. R. Nov 5 '15 at 12:38
• Only this much info is given in the question.The vector in the new position is given to be $\frac{4}{\sqrt2}\hat{i}-\frac{1}{\sqrt2}\hat{j}-\frac{1}{\sqrt2}\hat{k}$. – Vinod Kumar Punia Nov 5 '15 at 12:41
• Are you familiar with the finite rotation formula in 3D? :) – H. R. Nov 5 '15 at 12:47
• No,i dont know,what it is? – Vinod Kumar Punia Nov 5 '15 at 12:48

If we say that the vector moves in such a way that it touches the $x$ axis then we (perhaps) may assume that it moves in the plane determined by $\vec i$ and $\vec i+2\vec j+ 2\vec k$, the vector itself. A normal vector to this plane is $\vec i \times (\vec i+2\vec j+ 2\vec k)=-2\vec j+2 \vec k.$ The equation of such a plane is $y=z.$

So we have the following equations

$$x^2+2z^2=9$$ $$x+4z=0.$$

From here

$$z=\pm\frac{1}{\sqrt2}.$$

The rest follows. Note that there are two solutions in general. But one of them does not touch the $x$ axis while moving along the simplest path. Our (Okham) solution is

$$y=z=-\frac{1}{\sqrt{2}}\,\,\text{and }\,\, \, x=\frac4{\sqrt2}$$

in other words the new position of $\vec{OP}$ is $$\frac4{\sqrt2}\vec i-\frac{1}{\sqrt{2}}\vec j- \frac{1}{\sqrt{2}}\vec k.$$

Systematic Approach

Every rotation has two characteristics. The first one is the axis of rotation and the second one is the angle of rotation. Suppose that you have a given vector $\bf{r}$. Then, we want to find the rotation of this vector around the axis $L$ with the director $\bf{l}$ by the angle $\Phi$. If we call the vector after rotation $\bf{r'}$, then it can be proved that the following formula will hold

$$\boxed{ {\bf{r'}} = \cos \Phi {\bf{r}} + \sin \Phi {\bf{l}} \times {\bf{r}} + \left( {1 - \cos \Phi } \right)\left( {{\bf{r}}.{\bf{l}}} \right){\bf{l}} }$$

you can find the proof on the net. Now, in your question we have

\eqalign{ & {\bf{r}} = \left( {1,2,2} \right) \cr & \Phi = \frac{\pi }{2} \cr & {\bf{l}} = \frac{{{\bf{r}} \times {\bf{i}}}}{{\left\| {{\bf{r}} \times {\bf{i}}} \right\|}}\,\,\,\,\,\, \to \,\,\,\,\,{\bf{r}}.{\bf{l}} = 0 \cr}

put it into the formula and do the computations. In this case, the formula reduces to

$${\bf{r'}} = {\bf{l}} \times {\bf{r}}$$

If you want to solve the problem by writing down equations and solving for the three unknown components of ${{\bf{r'}}}$ there is some other ways. In your problem ${\bf{r'}}.{\bf{l}} = 0$ and $\Phi = \frac{\pi }{2}$ hold. We can take advantage of these. So the system that you should solve is
\eqalign{ & {\bf{r'}}.{\bf{r}} = 0 \cr & r' = r \cr & {\bf{r'}}.{\bf{l}} = 0 \cr}