Question

If the vertices A, B, C of a triangle ABC are (1, 2, 3), (-1, 0, 0), (0, 1, 2),respectively, then find ∠ABC. [∠ABC is the angle between the vectors $\overline{BA} \ and \ \overline{BC}$] .

Answer 1

Now, Vector AB = (-1 - 1)i + (0 - 2)j + (0 - 3)k

= -2i -2j -3k

Therefore, Vector BA = 2i + 2j + 3k

Similarly, Vector BC = (0 - (-1))i + (1 - 0)j + (2 - 0)k

= i + j + 2k

Now, it is given that Angle is in between BA and BC.

Therefore, for finding the angle, using the formula of dot product,

|A.B| = |A||B|Cosθ

Now, dot product of Vector BA with BC = (2 × 1) + (2 × 1) + (3 × 2)

= 2 + 2 + 6  = 10

Magnitude of BA = √(4 + 4 + 9) = √17

Magnitude of BC = √(1 + 1 + 4) = √6

∴ Cosθ = 10/(√17 × √6)

θ = Cos⁻¹(10/√102)

Hope it helps.