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