My prof in the class got $x=1+2s$, $y=1+2t$ and $z=3+s+2t$ but in the book says $x+2y-2z+3=0$. Both of them were first solving for $r(u)$ and $r(v)$ and I get that part but in the book they were using the cross product then and my prof wasn't, he was substituting $s$ and $t$. Can someone tell me what the correct answer is or if they are both correct?

  • 1
    $\begingroup$ Eliminate the parameters from $$x=1+2s,\;y=1+2t,\;z=3+s+2t$$ you get $s=(x-1)/2;\;t=(y-1)/2$ and plug in the third $z=3+(x-1)/2+y-1$ you have exactly the equation of the book $$x+2 y-2 z+3=0$$ $\endgroup$
    – Raffaele
    Nov 20, 2017 at 16:43

1 Answer 1


You have two options to write the equation of the tangent plane. It is the span of the two independent tangent vectors, so parametrically, it's $\mathbf{r}=\mathbf{r}_0+s\mathbf{r}_u+t\mathbf{r}_v.$ This is presumably what your prof did.

Or it can be written as an implicit equation in coordinates, then it's $\mathbf{n}\cdot(\mathbf{r}-\mathbf{r}_0)=0,$ where here the normal vector is $\mathbf{n}=\mathbf{r}_u\times\mathbf{r}_v.$ This is presumably what your textbook did.

Let's see them both in action. First we have $\mathbf{r}_u=(2u,0,1)$ and $\mathbf{r}_v=(0,2v,2).$ Evaluated at $(1,1,3),$ the tangent vectors are $(2,0,1)$ and $(0,2,2)$. So our tangent plane is $(x,y,z)=(1,1,3)+s(2,0,1)+t(0,2,2)=(1+2s,1+2t,3+s+2t).$


$$ \mathbf{r}_u\times \mathbf{r}_v=\det\begin{pmatrix}i & j & k\\2 & 0 & 1\\0 & 2 & 2\end{pmatrix}=(-2,-4,4). $$

Therefore the equation of our plane is $-2(x-1)-4(y-1)+4(z-3)=0,$ or $x+2y-2z+3=0.$

And we can check that the two sets of equations define the same plane by plugging $(x,y,z)=(1+2s,1+2t,3+s+2t)$ into $x+2y-2z+3$ and checking whether we get $0$.

  • $\begingroup$ Yeah, I totally get it now. Thank you very much, I appreciate your work!!! $\endgroup$
    – George S
    Nov 20, 2017 at 16:18
  • $\begingroup$ @DjordjeS thanks mate! $\endgroup$
    – ziggurism
    Nov 21, 2017 at 1:19

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