a) Dot product: recall that
i • i = j • j = k • k = 1
i • j = i • k = j • k = 0
Then
a • b = (i + 2j - 7k) • (5i - 2j + 4k)
a • b = 5 (i • i) - 4 (j • j) - 28 (k • k)
a • b = 5 - 4 - 28
a • b = -27
b) Cross product: recall that
i × i = j × j = k × k = 0
i × j = -(j × i) = k
j × k = -(k × j) = i
k × i = -(i × k) = j
Then
a × b = (i + 2j - 7k) × (5i - 2j + 4k)
a × b = -2 (i × j) + 4 (i × k) + 10 (j × i) + 8 (j × k) - 35 (k × i) + 14 (k × j)
a × b = -12 (i × j) - 6 (j × k) - 39 (k × i)
a × b = -6i - 39j - 12k
c) Recall the dot product identity
x • y = ||x|| ||y|| cos(θ)
We have
||a × b|| = √((-6)² + (-39)² + (-12)²) = 9√21
Then the direction cosines α, β, γ of a × b are
(a × b) • i = ||a × b|| ||i|| cos(α)
⇒ cos(α) = -6/(9√21) = -2/(3√21)
(a × b) • j = ||a × b|| ||j|| cos(β)
⇒ cos(β) = -39/(9√21)
(a × b) • k = ||a × b|| ||k|| cos(γ)
⇒ cos(γ) = -12/(9√21) = -4/(3√21)
d) a × b is perpendicular to both a and b, so we can get a unit vector by scaling this down by its magnitude. This is the same as the vector of direction cosines:
(a × b)/||a × b|| = (-6i - 39j - 12k)/(9√21)
(a × b)/||a × b|| = -2/(3√21) i - 39/(9√21) k - 4/(3√21) k
