Question

Suppose you need to know an equation of the tangent plane to a surface S at the point P (2,1,3). You don't have an equation for S but you know that the curves:

r1(t) = (2+3t, 1-t^2, 3-4t+t^2)


r2(t) = (1-u^2, 2u^3-1, 2u+1)


both lie on S. Find an equation of the tangent plane at P.


*Note: both r1 and r2 are vectors

Answer 1

24(x-2) -14(y-1) +18(z-3)=0

24x-14y+18z=88

Explanation:

Note:

There is a correction. + sign should be there in first term of r₂(u) because with - sign we can not solve the problem.

r₂(u)=(1+u^2, 2u^3-1, 2u+1)

We now that r₁(t) and r₂(u) both lie on the surface. So we will first find vectors B₁(t)=d/dt(r₁(t)) and B₂(u)=d/du(r₂(u))

B₁(t)=d/dt(2+3t ,1-t^2, 3-4t+t^2)

B₁(t)=(3, -2t, -4+2t)

B₂(u)=d/du(1+u^2, 2u^3-1, 2u+1)

B₂(u)=(2u, 6u^2, 2)

Put t=0 in r₁(t) we will get r₁(t)=(2, 1, 3) which is point P. So put t=0 in B₁(t) and we will get B₁(t)=(3,0,-4)

Similarly put u=1 in r₂(u) we will get r₂(u)=(2, 1, 3) which is point P. So put u=1 in B₂(u)=(2,6,2)

Take Cross Product of B₁(t) and B₂(u) as both are perpendicular to the surface

B₁(t) x B₂(u) =(3,0,-4) x (2,6,2)

B₁(t) x B₂(u) =(24, -14, 18)

So Equation of plane will be:

24(x-2) -14(y-1) +18(z-3)=0

24x-14y+18z=88