Michaelyayehidn
Answered

IDNLearn.com offers a comprehensive platform for finding and sharing knowledge. Get step-by-step guidance for all your technical questions from our dedicated community members.

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 :

LammettHashidn

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

Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your questions find clarity at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.