Question

the velocity of a particle is v vector =6i+2j-2k. the component of the velocity of a particle parallel to vector a vector =i+j+k.​

Answer 1

Answer:

Explanation:

"the velocity of a particle is v vector =$(6i+2j-2k)$

The component of the velocity of a particle parallel to vector a vector = i+j+k will be as follows:

v=$(6i+2j-2k)$

a= $(i+j+k)$

v.a= $(6i+2j-2k)$. $(i+j+k)$

v.a= 6×1+2×1−2×1

v.a= 6+2−2

v.a= 6

Magnitude of a= $(a)^{2}$= $(i+j+k)^{2}$=$(1^{2} +1^{2} +1^{2} )$ = 1+1+1 =3

Here $a^{2}$ =3 v.a= 6 a= (i+j+k)

Hence,

$(\frac{v.a}{a^{2} } ).a$= $(\frac{6}{3}) (i+j+k)$ =2 $(i+j+k)$

Component of the velocity parallel to the vector $(i+j+k)$ =2 $(i+j+k)$