Answer:
a. v.v' = v₁v₂ - v₁v₂ = 0 b. (20, -21)/29 and (-20,21)/29
Step-by-step explanation:
a. For two vectors a, b to be orthogonal, their dot product is zero. That is a.b = 0.
Given v = (v₁, v₂) = v₁i + v₂j and v' = (v₂, -v₁) = v₂i - v₁j, we need to show that v.v' = 0
So, v.v' = (v₁i + v₂j).(v₂i - v₁j)
= v₁i.v₂i + v₁i.(- v₁j) + v₂j.v₂i + v₂j.(- v₁j)
= v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j
i.i = 1, i.j = 0, j.i = 0 and j.j = 1
So, v.v' = v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j
= v₁v₂ × 1 - v₁v₁ × 0 + v₂v₂ × 0 - v₂v₁ × 1
= v₁v₂ - v₂v₁
= v₁v₂ - v₁v₂ = 0
So, v.v' = 0
b. Now a vector orthogonal to the vector v = (21,20) is v' = (20,-21).
So the first unit vector is thus a = v'/║v'║ = (20, -21)/√[20² + (-21)²] = (20, -21)/√[400 + 441] = (20, -21)/√841 = (20, -21)/29.
A unit vector perpendicular to a and parallel to v is b = (-21, -20)/29. Another unit vector perpendicular to b, parallel to a and perpendicular to v is thus a' = (-20,-(-21))/29 = (-20,21)/29
