0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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