InterviewSolution
Saved Bookmarks
| 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)` |
|