Find a decomposition of a=〈−3,4,−4〉a=〈−3,4,−4〉 into a vector cc parallel to b=〈−8,4,−8〉b=〈−8,4,−8〉 and a vector dd perpendicular to bb such that c+d=ac+d=a.

c=?

d=?

c=(-4,2,-4) and d= ( 1,2,0)

Step-by-step explanation:

in order to decompose a in 2 vectors c+d such that c+d=a , then we can find the scalar product of vectors a and b

a*b = |a|*|b|*cos(a,b)

but |a|*cos(a,b) = projection of a in b = modulus of a vector c parallel to b and decomposed from a = |c|

therefore

a*b = |b|*|c|

but also

a*b = ax*bx + ay*by + ac*bc = (-3)*(-8) + 4*4  + (-4)*(-8)= 72

|b| = √[(-8)² +4²+ (-8)²]= 12

then

|c|  =  (a*b)/|b| = 72/12 = 6

since c is parallel to b then  

c= |c|* Unit vector parallel to b =|c|* (b/|b|) = b * |c|/|b| = (−8,4,−8) * (6/12) = (-4,2,-4)

c=(-4,2,-4)

since

c+d=a → d=a-c = (−3,4,−4) - (-4,2,-4) = ( 1,2,0)

d= ( 1,2,0)

to verify it

d*b= dx*bx + dy*by + dc*bc = 1*(-8) + 2*4  + 0*(-8)= 0

therefore d is perpendicular to b as expected (the same could be verified with d*c)

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

no

step-by-step explanation: