Question

Tangent to the plane

Answer 1

Could someone help me on this?
Q 42

Answer 2

tangent vector of of the vector r1 and r2 can be gained by differentiating,
$$\mathbf{r_1}'(t)=\frac{d\,\mathbf{r}_1(t)}{dt}=<3,-2t,-4+2t>$$
$$\mathbf{r_2}'(u)=\frac{d\,\mathbf{r}_2(t)}{du}=<2u,6u^2,2>$$
at the point P(2,1,3), t=0 and u=1
$$\mathbf{r_1}'(0)=<3,0,-4>$$
$$\mathbf{r_2}'(1)=<2,6,2>$$
now we know two vectors parallel to the tangent plain and a point on the plain.

Vector normal to the plain = $$\mathbf{n}=\mathbf{r_1}\times\mathbf{r_2}$$

$$\mathbf{n}\cdot<2-x,1-y,3-z>=0$$

Answer 3

how would you find the equation of the tangent though?

Answer 4

by differentiating

Answer 5

actually you dont need the equation of the tangent. Just need two tangent vector