1.

Let A be a point on the line `vec(r)=(1-3mu)hati+(mu-1)hatj+(2+5mu)hatk` and `B(3,2,6)` be a point in the space. Then the value of `mu` for which the vector `vec(AB)` is parallel to the plane `x-4y+3z=1` is: (a) `1/8` (b) `1/2` (c) `1/4` (d) `-1/4`A. `(1)/(4)`B. `-(1)/(4)`C. `(1)/(8)`D. `(1)/(2)`

Answer» Correct Answer - A
Given equation of line is
`r=(1-3mu)hati+(mu-1)hatj+(2+5mu)hatk`
Clearly, any point on the above line is of the form `(1-3mu,mu-1,2+5mu)`
Let A be `(-3mu+1,mu-1,5mu+2)`
for some `muinR.`
Then, `AB(3-(-3mu+1))hati+(2-(mu-1))hatj`
`+(6-(5mu+2))hatk[becauseAB=OB-OA]`
`=(3mu+2)hati+(3-mu)hatj+(4-5mu)hatk" "...(i)`
Normal vector (n) of the plane `x-4y+3z=1` is
`n=hati-4hatj+3hatk" "...(ii)`
`because`AB is parallel to the plane.
`:.` n is perpendicular to the AB.
`impliesAB.n=0`
`implies[(3mu+2)hati+(3-mu)hatj+(4-5mu)hatk].[hati-4hatj+3hatk]=0`
[From Eqs. (i) and (ii)]
`implies(3mu+2)-4(3-mu)+3(4-5mu)=0`
`implies" "-8mu+2=0`
`implies" "mu=(1)/(4)`


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