Question

Plane equation for N = < e , -pi/2 , 2 > / sqrt( e^2 + pi^2/4 + 4) ) and P = ( e, 0 , 1 )

Answer 1

can't I just use the vector in the numerator?

Answer 2

don't understand the problem, N is normal vector of what? the curve? the plane? P is point?

Answer 3

equation I got is ex - pi/2 y + 2z = e^2 + 2

Answer 4

N is the normal vector of the plane
P is the point

Answer 5

your N is unit normal vector, or just normal? I think it's unit since you divided by magnitude of it, right?

Answer 6

so, you just pick the numerator of that unit to form the form of plane?

Answer 7

that magnitude is in the denominator...

Answer 8

nope, show me your work

Answer 9

so the actual problem is

Answer 10

Find the equation of the plane normal to r(t) = < e^t sin( (pi/2) t ) , e^t cos( (pi/2) t), t^2 > when t = 1

Answer 11

I calculatee r'(t)

Answer 12

I'll take a picture of my work

Answer 13

I think you just put t =1 to the normal vector, to get it value, no need to take r'(t) because r'(t) used to find tangent plane at the point. not plane which has that normal vector.

Answer 14

the normal plane

Answer 15

has the normal vector that is the Tangent vector of r(t)

Answer 16

you confused,