Passes through the point $(3,4,5)$ and contains the line $x=5t,$ $y=3+t$ $,z=4-t$.

To start out with the parametric equations really give a point and a direction vector with the point being (0,3,4) Now if I take the cross product of two points (also vectors with tails at the origin) I get the normal vector of the plane containing these two points:

$-i+12j-9k \space$ this is the normal vector I get when doing the determinant of the $(3,4,5)$ and $(0,3,4)$. Then of course I take my normal vector and multiply it by $(x,y,z)-(3,4,5)$ which gave me $-1(x-3)+12(y-4)-9(z-5)=0$ but apparently this is incorrect I am not sure why.

  • $\begingroup$ Are you sure that this vector is orthogonal to the plane? $\endgroup$ – user178826 Oct 12 '16 at 21:42
  • $\begingroup$ if and only if the dot product of it and any point in the plane is 0 i believe so: (-1)(3)+4(12)-9(5)= -3+48-45=0 so yes $\endgroup$ – K. Gibson Oct 12 '16 at 21:47
  • $\begingroup$ Is the equation of plane $x-4y+z+8=0$ $\endgroup$ – Navin Oct 12 '16 at 21:53
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    $\begingroup$ this must happen not for any point, but for any vector determined by two points of the plane. I mean if n is orthonormal and A and B are ooints of the plane, we need $n\cdot\overline{AB}=0$ $\endgroup$ – user178826 Oct 12 '16 at 21:57

the vector $-i+12j-9k$ is not orthogonal to the plane. Instead, you can consider the director vector of the line $5i + j - k$ and de vector from $(3,4,5)$ to $(0,3,4)$, and then you can obtain the orthonormal vector with the determinant.

  • $\begingroup$ but if it dots to zero how is not orthogonal to the plane? $\endgroup$ – K. Gibson Oct 12 '16 at 21:51

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