Name:TEST 1answersNo Calculators1:2:3:4:5:1 (48 points) A hummingbird starts at a feeder located at F (0, 0, 10) (distances infeet) and flies straight toward a point B(10, 20, 30) on a branch with a spee d 3ft/s. Anobserver is located at O(5, 10, 10).(a) (5) How long does it take the hummingbird to reach the branch?−−→F B =< 10, 20, 20 >, |−−→F B| =√100 + 400 + 400 = 30travel time =|−−→F B|speed=303= 10 seconds.(b) (5) What is the hummingbird’s velocity vector?v = speed × unit direction vector = 31|−−→F B|−−→F B =< 1, 2, 2 >(c) (5) Write down the equation of the line that contains the hummingbird’s path.r = r0+ vt where r0=< 0, 0, 10 > is the starting point, hencex = t y = 2t z = 10 + 2t(d) (5) Where is the hummingbird going to be in 2 s ec onds?x = 2 y = 2 ∗ 2 = 4 z = 10 + 2 ∗ 2 = 14e) (7) What is the projection of−−→F O onto−−→F B?proj−−→F B−−→F O =−−→F B ·−−→F O|−−→F B|2−−→F B =259< 1, 2, 2 >=259,509,509f) (7) What is the equation of the plane that contains the triangle 4F OB?−−→F B ×−−→F O =i j k10 20 205 10 0=< −200, 100, 0 >= normal to the plane(0, 0, 10) is a point in the plane, hence−200(x − 0) + 100(y −0) + 0(z − 10) = 0 or 2x − y = 0g) (7) What is the area of the triangle 4F OB?area =12|−−→F B ×−−→F O| =12p2002+ 1002= 50√5h) (7) How close to the observer will the hummingbird get?using g:distance =|−−→F B ×−−→F O||−−→F B|=100√530using e:distance =−−→F O −proj−−→F B−−→F O=s5 −2592+10 −5092+5092=10√532 (24 points)a) Find the parametric equation of the line through the point P (1, 0, −1) and per-pendicular to the plane 5x − 2y + 3z = 7.normal to the plane = v =< 5, −2, 3 >line: x = 1 + 5t y = 0 − 2t z = −1 + 3tb) Find the point R where this line intersects the plane.finding t such that 5(1 + 5t) − 2(−2t) + 3(−1 + 3t) = 7 gives t =538x = 1 +2538=6338y = −1038z = −1 +1538= −2338R =6338, −1038, −2338c) Find the distance of the point P from the plane.using bdistance = |−→P R| =s1 −63382+10382+−1 +23382=5√38as in the book, p.886, choos ing a point Q = (1, −1, 0) on the plane givesdistance =|−−→P Q · v||v|=5√383 (18 points) Find an equation of the plane that contains the origin and the linex = 1 + 2t, y = 1 − t, z = 2t, −∞ < t < ∞.vector parallel to the line v =< 2, −1, 2 >point on the line ( t = 0): Q = (1, 1, 0), position vector u =< 1, 1, 0 >.normal to the plane:v × u =i j k2 −1 21 1 0=< −2, 2, 3 >plane: −2x + 2y + 3z = 04 (10 points) Sketch and identify the surface x2+ y2− z2+ 1 = 0.In the plane z = c we have a circle x2+ y2= c2− 1radius√c2− 1, hence |z| has to be ≥ 1.So, we have hyperboloid of two sheets (p. 895)5 (10 extra credit points if you get 90 or more points on problems 1-4 points) Find theprojection of the liner = r0+ tv, −∞ < t < ∞onto a plane which contains a point with a position vector r1and has a normal n. Hint:Start by following the first steps in problem 2.Let P (t) be a point at a fixed t on the line. The equation of the line perpendicularto the plane and passing through P (t) isr = r0+ tv + sn, −∞ < s < ∞.It intersects the plane when(r0+ tv + sn − r1) · n = 0hences = −(r0+ tv − r1) · n|n|2.Hence the projection of P (t) on the plane isr0+ tv −(r0+ tv − r1) · n|n|2n- which represents a line with a directionv −v · n|n|2n = v − projnvand passing through a pointr0−(r0− r1) ·