At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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
The normal plane would include the vector perpendicular to that vector if I remember right
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