Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I have a given point $A(x_a,y_a,z_a)$ in the horizontal plane $z=1400$ and I rotate the plane in such a way that it is perpendicular to a line that makes an angle of $60^{\circ}$ with $Oz$ axes and its projection makes $45^{\circ}$ and $45^{\circ}$ angles with $Ox$ and $Oy$, what will the new coordinates of the point be?

Here's what I managed to do so far:

I found the vector perpendicular to the plane $$\frac{3\sqrt{2}}{4\sqrt{2}} i + \frac{3\sqrt{2}}{4\sqrt{2}} j + \frac{1}{2}k;$$ the new plane contains the point $(x_a,y_a,z_a)$, so the new plane equation is: $$\frac{3\sqrt{2}}{4\sqrt{2}}(x-x_a)+\frac{3\sqrt{2}}{4 \sqrt{2}} (y-y_a)+\frac{1}{2}(z-z_a)=0.$$ I have one equation missing (since $x_{a,new}=x_a$).

Thanks

share|improve this question
1  
Could you try to explain what you have tried so far and where you are stuck? –  t.b. May 9 '11 at 11:14
    
I found the vector perpendicular to the plane: 3*sqrt(2)/(4*sqrt(2))i+3*sqrt(2)/(4*sqrt(2))j+0.5k; the new plane contains point (xa,ya,za) so the new plane eq. is: 3*sqrt(2)/(4*sqrt(2))(x-xa)+3*sqrt(2)/(4*sqrt(2))(y-ya)+0.5(z-za)=0. I have one eq. missing (since xa_new=xa) .... –  amu May 9 '11 at 15:04
    
Am I right to assume that the x-axis in the old plane corresponds to the $\displaystyle \frac{3\sqrt{2}}{4\sqrt{2}}i + \frac{3\sqrt{2}}{4\sqrt{2}}j - \frac{\sqrt{3}}{2}k$ vector in the new plane? –  El'endia Starman May 9 '11 at 23:42

1 Answer 1

If it's not too late to offer help on this matter, I'd like to offer how I'd solve the problem. Seeing you already calculated the vector normal to second plane, let's call it $\overrightarrow{n_2}$ and vector $(0,0,1)$ perpendicular to first plane $\overrightarrow{n_1}$... The angle of rotation will be $cos(\alpha)=\frac{\overrightarrow{n_1}\cdot\overrightarrow{n_2}}{|\overrightarrow{n_1}|\cdot|\overrightarrow{n_2}|}$ , and the axis vector (for rotation) will be $\overrightarrow{N}=\frac{\overrightarrow{n_1}\times\overrightarrow{n_2}}{|\overrightarrow{n_1}\times\overrightarrow{n_2}|}$. All you have to do then is to rotate the point represented by vector $\overrightarrow{A}$ using the formula: $\overrightarrow{A_{new}}=(\overrightarrow{N}\cdot\overrightarrow{A})\cdot\overrightarrow{N}+sin(\alpha)\cdot(\overrightarrow{N}\times\overrightarrow{A})+cos(\alpha)\cdot(\overrightarrow{N}\times\overrightarrow{A})\times\overrightarrow{N}$ .

Hope it helps...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.