Question

In the diagram, OAB and OED are straight lines.
O is the origin, A is the midpoint of OB and E is the midpoint of AC.
AC = a and CB = b.
Q.) Find vector OD.​

Answer 1

Step-by-step explanation:

if the tangent to the curve y= x^3 +ax+b at (1,6) is parallel to the line x-y+5=0 find a and b

Answer 2

Answer:

CD = mb

DB = nb

CD + DB = mb + nb = (m+n)b

CB = CD + DB

b = (m + n)b

=> m + n = 1     (1)

OA = AB = a +b

OB = OA + AB = 2a + 2b

OE = OA + AE = a + b + a/2 = (3/2)a + b

OD = p OE = (3p/2)a + pb   (2)

OD = OB + BD = OB + -DB = 2a + 2b -nb = 2a + (2-n) b (3)

from 2 & 3

(3p/2)a + pb = 2a + (2-n) b

equating coefficients

(3p/2) = 2 or p = 4/3

p =2-n  or n = 2 -p  

n = 2 - 4/3 =2/3

so m = 1 - n = 1 - 2/3 = 1/3

from 3

OD = 2a + (2-n) b

= 2a + (2 - (2/3) )b

=2a + (4/3)b

Step-by-step explanation: