Question

Find tha angle between the vectors:i^-j^ and j^-k^

Answer 1

Answer: cos⁻¹((-1) / 2)=2.0943951023932

Step-by-step explanation:

A=i^-j^ ,B=j^-k^

A.B=|A|x|B|xcos(X))

A.B/|A|×|B|=cos(X)

X= cos⁻¹(A.B/|A|×|B|)

X=cos⁻¹(-1/((sqrt(2) sqrt(2))

X=cos⁻¹((-1) / 2)

X=2.0943951023932

Answer 2

Answer:

120°

Step-by-step explanation:

• let vectors be a and b.
• take dot product ( 1i - 1j + 0k).(0i + 1j + 1k)
• which gives (1)(0)+(1)(-1)+(0)(1) = -1
• find magnitude of both vectors
• both have magnitude sqrt(1²+1²)=sqrt(2)
• a.b = |a| |b| cos(θ)
• put the values
• -1 = sqrt(2) sqrt(2) cos(θ)
• -1 = 2 cos(θ)
• cos(θ)= -1/2
• θ = cos⁻¹(-1/2) = 120°