Calculus: Find the equations of the tangent line and normal plane to the curve R(t) = t^2i + t^3j + tk at t=1

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Calculus: Find the equations of the tangent line and normal plane to the curve R(t) = t^2i + t^3j + tk at t=1

Mathematics
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R(1) = 1i +1j +1k; so it is comprised of unit vectors
R' = 2ti + 3t^2j +k
the slope of at t=1 should be parallel to the vector <2,3,1> i think

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The normal plane would include the vector perpendicular to that vector if I remember right
dR/dt=2ti+3t^2j+k
what's that equation for?
R' is the slope of the tangemt to the curve; to get the equation of it you have to include it as the slope of the line right?
R' is normal to the tangent plane.
normal as in perpendicular?
So the form of the tangent plane will be\[2x+3y+z=d\]and since the point (1,1,1) lies in it (where t=1), you have d=6, so\[2x+3y+z=6\]
Normal plane, I meant.
The velocity vector is tangent to the curve, so it's perpendicular to the normal plane. That's what I mean to say above.
Thanks so much, what would be the tangent line then?
tangent line is (1,1,1)+t(2,3,1)
Anchor point + tangent vector x parameter...like uzma's done.
there are many tangent lines to the curve at that point. they make up the plane :)
but here normal plane is required
like this right?
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