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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.

Please help me.Thanks.

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  • $\begingroup$ "turns through a right angle" about which axis??? or which vector???? Info required. $\endgroup$ – SchrodingersCat Nov 5 '15 at 12:38
  • $\begingroup$ What is the axis of rotation? :) $\endgroup$ – H. R. Nov 5 '15 at 12:38
  • $\begingroup$ Only this much info is given in the question.The vector in the new position is given to be $\frac{4}{\sqrt2}\hat{i}-\frac{1}{\sqrt2}\hat{j}-\frac{1}{\sqrt2}\hat{k}$. $\endgroup$ – Vinod Kumar Punia Nov 5 '15 at 12:41
  • $\begingroup$ Are you familiar with the finite rotation formula in 3D? :) $\endgroup$ – H. R. Nov 5 '15 at 12:47
  • $\begingroup$ No,i dont know,what it is? $\endgroup$ – Vinod Kumar Punia Nov 5 '15 at 12:48
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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 $$\frac4{\sqrt2}\vec i-\frac{1}{\sqrt{2}}\vec j- \frac{1}{\sqrt{2}}\vec k.$$

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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

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


Your Approach

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

$$\eqalign{ & {\bf{r'}}.{\bf{r}} = 0 \cr & r' = r \cr & {\bf{r'}}.{\bf{l}} = 0 \cr} $$

where the first one states that the rotated vector is perpendicular to the original one. The second one says that the length of the rotated vector and the original vector are equal. The third one emphasize this fact that the rotated vector is in the plane perpendicular to the rotation axis.

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