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Find the point of intersection of the tangent plane to the surface $z+1=xe^y\cos(z)$ at the point $(1,0,0)$ and the line $L$ given by: $x=2t, y=t+1, z=1-3t$.

Paul
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user177776
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  • Have you figured out the equation for the tangent plane yet? Hint: There is a formula for that in your textbook! – Jyrki Lahtonen Sep 22 '14 at 07:04
  • equation would be the dot product and I need up with e^ycos(z) – user177776 Sep 22 '14 at 15:56
  • Hmm. Are you trying to solve for $z$? That won't work this time. Write the equation of the surface in the form $$F(x,y,z)=0$$ and use the fact that the gradient $\nabla F$ (evaluated at the point of interest) is a normal to the tangent plane. – Jyrki Lahtonen Sep 22 '14 at 16:07
  • Okay, so would the equation be x+y-z-1? – user177776 Sep 22 '14 at 16:28
  • That seems to be the case. Well done! Do you know how to find the value of $t$ that corresponds to the point of intersection? – Jyrki Lahtonen Sep 22 '14 at 16:34
  • No, this is where I'm confused at, we never went over this part in class – user177776 Sep 22 '14 at 16:35
  • That may have been explained in an earlier class (in these parts finding the point of intersection of a plane and a parametrized line is done in Linear Algebra). Here's how: substitute the parametric forms of $x,y,z$ into the equation of the tangent plane, and see what you can do with it. – Jyrki Lahtonen Sep 22 '14 at 16:37
  • so it would look like 2t+(t+1)-(1-3t)-1? then what? Don't you solve for t? but doesn't the equation need to be = to something or is it just 0? – user177776 Sep 22 '14 at 16:41
  • Oops, we forgot that from the equation of the tangent plane. Anyway, your guess is correct. The equation is $x+y-z-1=0$ :-) – Jyrki Lahtonen Sep 22 '14 at 16:43
  • See. It wasn't so difficult! – Jyrki Lahtonen Sep 22 '14 at 16:52

1 Answers1

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To specify the tangent plane we need a point on it (given) and its normal. This time the surface is given implicitly in the form $F(x,y,z)=0$ with $$ F(x,y,z)=x e^y\cos z-z-1. $$ The normal is then given by the gradient $$ \nabla F=(e^y\cos z, e^y x \cos z,-1-e^yx\sin z), $$ which evaluated at $P=(1,0,0)$ gives us the normal vector $$ \vec{n}=\nabla F(P)=(1,1,-1). $$ Thus the equation of the tangent $T$ is of the form $x+y-z=a$. The constant $a$ can be solved using the fact that $P\in T$. We get $a=1$. Substituting the parametrizations into the equation of the tangent gives us the equation $$ (2t)+(1+t)-(1-3t)=1. $$ From which we solve $t=1/6$. The sought after point of intersection is thus $$Q=(2\cdot\dfrac16,1+\dfrac16,1-3\cdot\dfrac16)=(1/3,7/6,1/2).$$

Here's a Mathematica pic of the surface (the blue thingy with mesh lines), the tangent plane (the semi-opaque grey plane), the line (red) and the point $Q$.

enter image description here

Jyrki Lahtonen
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