A particle is moving with speed 6 m/s along the direction of Ä = 2i+ 2j- K, then its velocity is :
(A) (4î+ 21 - 4k) units
(B) (41 +4j-2) units
(C) (4 + 4j-4K) units
(D) (2 + 4 -2K) units
and please provide solution!
Answers
Answer:
B
Explanation:
speed of particle is 6m/s
unit vector in the direction of Ä = ( 2i + 2j - k ) is
 = 2i + 2j - k / √ 4+4+1
 = 2i + 2j -k / 3
velocity of the particle is
= 6(2i +2j -k ) / 3
= 4i + 4j - 2k
Answer:
B
Explanation:
here is your answer .
given -
speed =6m/s
Position vector(A vector )= 2 i cap+2 j cap -K cap
position vector just shows the co-ordinates of the point in three dimensional space
remember unit vector is a vector dhowing directon only
Velocity= speed * direction
speed 6 m/s
direction = unit vector
A cap is simple to calculate
Observe
from formula
=6 * (2 i cap+2 j cap-k)/3
now if your are wondering what is unit vector beside 6
simply calculate unit vector of A vector
=4 i cap + 4 j cap - 2 k cap
option B is correct .
fact is - i had the test in that this question was the first
ok dont mark me the brainliest