Find all the points P = P(x, y, z) on the Sphere: S: x^2 + y^2 + z^2 = 4, that a

Question

Find all the points P = P(x, y, z) on the Sphere: S: x^2 + y^2 + z^2 = 4, that are closest to. and farthest from the point Q = Q(3, 1, -1) in R^3. (b) Find the tangential component a_T, and the normal component a_N of the acceleration vector a vector for the position vector: r vector(t) = ti vector + t^2j vector + e^xk vector, for t Element R. Then without computing the unit tangent vector T vector = T vector(t), and the unit normal vector N vector = N vector(t), write the acceleration vector a vector in the form: a vector = a_T T vector + a_N N vector, at t = 1.

Explanation / Answer

a.) the center is (0,0,0) and radius= 2 unit

Now according to question the point given is=(3,1,-1)

distance between center to given points=(11)0.5

so we get that = 110.5 > 2 it tells that point is outside sphere.

again assume closest point (X1,Y1,Z1) and farthest point (X2,Y2,Z2)

So this point will satisfy few condition-

X21 +Y12+Z21=4 .............(1)

X22+Y22+Z22=4 ...............(2)

X1+X2=0 , Y1+Y2=0 ,Z1+Z2=0 (In vector form).........(3)

SQURE ROOT [(X2-3)2+(Y2-1)2+(Z2+1)2] - SQURE ROOT [(X1-3)2+(Y1-1)2+(Z1+1)2] =2*RADIUS

Now find UNKNOWNS.

b.) tangent will be-

double differentiate w.r.t 't'

doing like that we get- aT =(0,2,et) =2j+etk (tengential component)

unit tengent vector= tengent/magnitude

=(2j+etk)/sqt(4+e2t)

normal vector(aN) = a - aT.

unit normal = aN / sqt(aN).

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