Find the distance of the point (3,4,-4) from the line r(t) = (1 + 3t, -1...
(1 point) Find the distance of the point (2,4, -4) from the line r(t) = (1 + 2t, 1 + 4t, 3 – 3t). Answer:
(1 point) Starting from the point (4,3,2)(4,3,2) reparametrize the curve r(t)=(4+3t)i+(3−3t)j+(2−2t)kr(t)=(4+3t)i+(3−3t)j+(2−2t)k in terms of arclength.
(1 point) Starting from the point (-4,-1,0) reparametrize the curve r(t) = (-4+ 3t)i + (-1+2t)j + (0+2t)k in terms of arclength. r(t(s)) it j+ k
please answer both (12(8 pts) Find parametric equations of the line through the point (2, -1,3) and perpendicular to the line with parametric equations 1-t,y 4- 2t and 3+ t and perpendicular to the line with parametric equations 3+t,y 2-t and z 3+2t. (13)(8 pts) Find the unit tangent vector (T(t) for the vector function r(t) - costi+3t j+ 2sin 2t k at the point where t 0 (12(8 pts) Find parametric equations of the line through the point (2,...
2. Find distance from point S(-2, 3, 4) to the line x = 3 - 2t, y = –2 + 3t, 2 = 5 - 6t Write plane equation passing through point S and par- allel to the given line. Show calculation steps clear and cleanly.
(1 point) Given R(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tkR(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tk Find the derivative R′(t)R′(t) and norm of the derivative. R′(t)=R′(t)= ∥R′(t)∥=‖R′(t)‖= Then find the unit tangent vector T(t)T(t) and the principal unit normal vector N(t)N(t) T(t)=T(t)= N(t)=N(t)= (1 point) Given R(t) = cos(36) i + e sin(3t) 3 + 3e"k Find the derivative R') and norm of the derivative. R'(t) = R' (t) Then find the unit tangent vector T(t) and the principal unit normal vector N() T(0) N() Note: Yn can can on the hom
4. Find the parametric equations for a line through a point (0,1,2) that (a). parallel to the plane x + y + z = 2, and (b). perpendicular to the line T = 1+t, y = 1 –t, z = 2t (Answer: x = 3t, y=1-t, 2 = 2 - 2t)
(1 point) Suppose a line is given parametrically by the equation L(t) = (3,2 – 3t, -2t) Then the vector and point that were used to define this line were U II , and p =
Find the equation of the plane through the point (-2,8,10) and parallel to the line x=1+t, y=2t, z=4-3t
(1 point) Find the solution r(t) of the differential equation with the given initial condition: r' (t) = (sin 2t, sin 5t, 3t) , r(0) = (5,8,6) r(t) =(