Question

# Velocity of a particle is →v=6^i+2^j−2^k. The component of the velocity parallel to vector a=^i+^j−^k in vector form is6^i+2^j+2^k^i+^j+^k2^i+2^j+2^k6^i+2^j−2^k

Solution

## The correct option is C 2^i+2^j+2^kGiven, a vector →v  and a unit vector ^n, you can divide the vector →v into two parts →v=→p+→q where  →p and →q  are parallel and perpendicular to ^n respectively. Now, the magnitude of →p is given by →p.^n . Since →q.^n=0, we have, |→p|=→p.^n=→v.^n thus,  →p=|→p|^n=(→v.^n)^n Since the unit vector in the direction of →a is given by →a|→a|, the component of →v parallel to →a is given by  (→v.→a|→a|)→a|→a|=(→v.→a)→a|→a|2 So, for the given problem, the component that we get is (6^i+2^j−2^k)(^i+^j+^k)(^i+^j+^k)(^i+^j+^k)(^i+^j+^k)=2^i+2^j+2^k

Suggest corrections