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)