Given the vectors a=3i+4j, b=-2i+5j, c=10i-j, d=(-1/3)i+(5/2)j, find: ǁ-10d-10c-10bǁ

Question

Given the vectors a=3i+4j, b=-2i+5j, c=10i-j, d=(-1/3)i+(5/2)j, find: ǁ-10d-10c-10bǁ

Answer 1

-10d = -10((-1/3)i + (5/2)j) = 10/3i + -25j

-10c = -10(10i - j) = -100i + 10j

-10b = -10)(-2i + 5j) = 20i + -50j

ǁ-10d-10c-10bǁ = ǁ-10d + -10c + -10b ǁ =

||10/3i + -25j + -100i + 10j + 20i + -50j || =

||(10/3i + -300/3i + 60/3i) + (-25j + 10j + -50j) || =

|| -230/3i + -65j || =

√(-230/3)² + (-65)²) =

√(52900/9 + 4225) =

√(52900/9 + 38025/9) =

√(90925/9) =

5/3√3637 ≈ 100.513

Answer 2

Given the vectors a=3i+4j, b=-2i+5j, c=10i-j, d=(-1/3)i+(5/2)j.

-10d = (10/3)i - 25 j

-10c = -100i + 10j

-10b = 20i - 50j

Therefore:

-10d-10c-10b = (10/3)i - 25 j + (-100)i + 10j + 20i - 50j

= (-230/3)i - 65j

ǁ-10d-10c-10bǁ = [(-230/3)² + (-65)²]^1/2 ≈ 100.51