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.
Best Answer
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 $$\frac4{\sqrt2}\vec i-\frac{1}{\sqrt{2}}\vec j- \frac{1}{\sqrt{2}}\vec k.$$