Question

cal 3 question. thanks and wish you the best ?

3. Find the equation of the plane that passes through (2, 1, 1), (1,4, 1) and (-2, 0,4). Put the equation in simplest general form. 4. Evaluate the velocity vector and acceleration vector of the object at the given value of t. r(t) = ti - tj + 9 - tak, t = 0

Answer 1

Griven points are: P(2,1, 961410) SR (-2,0,4) General eqn of the plane passing these through (xo, Yo, Zo) is given by; a(x-Xo) + bly-Yo) + c(z-zo) = 0 where <a,bicy is perpendicular to the plane Let J = P = <-1,3,0) VL-PR-4,-1,3) i } R V x 2 = - T (9)-J (-3) +R (1+12) 3 -4 3 v xiz 91 +37 +13 K =<9,3,137 = {abic) so 3 so af x-xo) tbly-yo) + C(Z-Zo)=0 9 (X-Xo) + 3(y-yo) +13 (Z - Z)=0 since passes through P(2,111) 9 (2-2) +314-1 +13(2-1) = 0 → 9x + 3y + 13 Z - 18-3-13-0 → 9x+3y +137 = 34 Hence General equation of the plane passes through (2,111), (1,411) & (-2,014) is. 9x+3y +13Z - 34 -Ans.

4 Given, ð t =ti - ti+ Ug-tz R VEU = 8(t) = 2 - 3 + at *=0 7 31(0)= =e-3 -t Jo-t² I a t R = ?-] tok so velocity ver for at tao is: Vio)= - j =<1,-1,0) ñ"t) = 07-01 39-42 (1) Now i tjen 3-EL 9t2 at t=0 g" (0) = oi-oj 59 - 0 TV 9-0 - = - tak - ž k 3 So at t=0 is: acceleration vector - 1 / 3 ã (o) = TV - <0,0,) Ans.