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2) Let a =i+2j-7k and b=5i-2j+4K Find
a) a.b
b) axb
C) The direction cosine of axb
D) The Unit Vector Perpendicular to both a&b​

Sagot :

LammettHash

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

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