Question

If i+ j, j+ k, i+ k are position vectors of vertices of triangle ABC taken in order, then ∠ A is equal to:

• A. \(\dfrac{\pi}{2}\)
• B. \(\dfrac{\pi}{5}\)
• C. \(\dfrac{\pi}{6}\)
• D. (π)/(3)

Hint:

Angle between vectors uses dot product.

Answer 1

The Correct Option is D

Step 1: vecAB=( j+ k)-( i+ j)= - i+ k.
Step 2: vecAC=( i+ k)-( i+ j)= - j+ k.
Step 3: cos A=fracvecAB·vecAC|vecAB||vecAC|=(1)/(2)⟹ A=(π)/(3).