Vector angle help?

Question

ABC is a triangle with vector AB= (i+2j+2k) and vector AC= (2i-j-k). Determine the largest angle in the triangle.

Answer 1

In triangle ABC, the vector AB = i + 2j + 2k and vector AC = 2i - j - k.

Find remaing vector BC = AC - AB = (2i - j - k) - (i + 2j + 2k) = i - 3j - 3k.

Length = |AB| = √(1² + 2² + 2²) = √9 = 3

Length = |AC| = √(2² + (- 1)² + (- 1)²) = √6 ≅ 2.45

Length = |BC| = √(1² + (- 3)² + (- 3)²) = √19 ≅ 4.36

From the above, BC is the largest side; hence the angle opposite to the longest side of a triangle is the greatest angle.

Hence here <A is the greates.

==> cos(A) = (AB.AC)/{|AB|*|AC|} = [ (i + 2j + 2k) * (2i - j - k) ] / [ 3*√6 ] = (2 - 2 - 2)/{3*√6) = - 2/(3√6) = - √6/9.

Since it is negative, angle A is obtuse.

A = cos^{- 1}(- √6/9) ≅ cos^{- 1}(- 0.27) ≅ 105.8°.