A uni-modular tangent vector on the curve `x=t^2+2,y=4t-5,z=2t^2-6t=2` is a. `1/3(2 hat i+2 hat j+ hat k)` b. `1/3( hat i- hat j- hat k)` c. `1/6(2 hat i+ hat j+ hat k)` d. `2/3( hat i+ hat j+ hat k)`

Options

A. `(1)/(3)(2hati+ 2hatj+hatk)`
B. `(1)/(3) (hati-hatj-hatk)`
C. `(1)/(6)(2hati + hatj + hatk)`
D. `(2)/(3)(hati +hatj +hatk)`

Answer

A

Explanation

The position vector of any point at `t` is
`vecr = (2+ t^(2) ) hati + (4t -5) hatj + ( 2t^(2) -6) hatk`
`rArr (dvecr)/( dt) = 2t hai + 4hatj + ( 4t -6) hatk`
`rArr " "(dvecr)/(dt):|_(t=2) = 4hati +4hatj+2hatk`
and `" "|(dvecr)/(dt)| :|_(t=2) = sqrt(16+16+4)=4`
Hence, the required unit tangent vector at `t = 2` is
`(1)/(3) ( 2hati + 2hatj +hatk)`.