Problem 11 - Entrance Test

Find the angle between the vectors a = i + 2j + 3k and b = 2i - j + k.

A. cos⁻¹(0)B. cos⁻¹(1/√14)C. cos⁻¹(1)D. cos⁻¹(√14)

Correct: B

Let θ be the angle between the vectors a and b. Then, cos θ = (a.b) / (|a||b|). a.b = (1)(2) + (2)(-1) + (3)(1) = 2 - 2 + 3 = 3. |a| = √(1² + 2² + 3²) = √(1 + 4 + 9) = √14. |b| = √(2² + (-1)² + 1²) = √(4 + 1 + 1) = √6. cos θ = 3 / (√14 * √6) = 3 / √84 = 3 / (2√21) = 3√21 / (2*21) = √21 / 14 = √3 * √7 / (2 * 7) = 1/sqrt(14*6)*3 = 3/sqrt(84)= 3/sqrt(4*21)=3/(2sqrt(21))=3sqrt(21)/(2*21) = sqrt(21)/14 = (sqrt(3*7))/14 = (sqrt(3)*sqrt(7))/(2*7). cos(θ) = (1*2 + 2*-1 + 3*1)/(sqrt(1+4+9)*sqrt(4+1+1)) = (2-2+3)/(sqrt(14)*sqrt(6)) = 3/sqrt(84) = 3/sqrt(4*21) = 3/(2sqrt(21))= (3sqrt(21))/(2*21)=sqrt(21)/14. Incorrect question and answer. cos(0) = 1