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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

  • 1
    Could you try to explain what you have tried so far and where you are stuck?2011-05-09
  • 0
    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) ....2011-05-09
  • 0
    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?2011-05-09

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