Question

A force of magnitude 50 KN is acting along the line joining A(2,0,6) and B(3,-2,0)m. Write the vector form of the force.​

Answer 1

Explanation:

The force acts along the line joining point A whose position vector OA is OA = 2 i + 0 j + 6 k) and point B with position vector OB = 3 i - 2 j + 0 k.

(The position vector of a point P is the vector directed from the origin to the point P and is written in bold face and is just the coordinates of the point with the unit vectors i, j, and k).

The vector directed from initial position A to final position B = A B = OB - OA = ( 3 i - 2 j + 0 k) - (2 i + 0 j + 6 k) = (3 - 2) i + (- 2 - 0) j + (0 - 6) k = i - 2 j - 6 k.

|A B| = [(+1)²:+ (-2)² + (-6)²]½ = (1 + 4 + 36)½= √41

So unit vector along AB = (1/√41)( i — 2 j — 6 k )