Let `veca = 2hati+(lambda)_1hatj+3hatk` , `vec(b)=4hati+(3-(labda)_2)hatj+6hatk` and `vec(c)=3hati+6hatj+((lambda)_3-1)hatk` be three vectors such that `vec(b)=2 vec(a)` and `vec(a)` is perpendicular to `vec(c)` then a possible value of `((lambda)_1,(lambda)_2,(lambda)_3)` is: (a) `(1,3,1)` (b) `((-1/2),4,0)` (c) `(1,5,1)` (d) `((1/2), 4, -2)`

Options

A. `(1,3,1)`
B. `(1,5,1)`
C. `(-(1)/(2), 4,0)`
D. `((1)/(2) , 4, -2)`

Answer

C

Explanation

We have `a= 2hat(i) + lambda_(1) hat(j) + 3hat(k) , b=4hat(i) + (3-lambda_(2)) hat(j) + 6 hat(k)`
and ` c= 3hat(i) + 6hat(j) + (lambda_(3)-1)hat(k)`
such that `b=2a`
Now ` b=2a`
` rArr 4hat(i) +(3 - lambda_(2)) hat(j)+ 6hat(k) =2 (2hat(i) + lambda_(1) hat(j) +3hat(k))`
` rArr 4hat(i) +(3 - lambda_(2)) hat(j)+6hat(k) =4 hat(i) +2lambda_(1) hat(j) +6hat(k)`
`rArr (3-2lambda_(1)) -lambda_(2))hat(j)=0`
`rArr 2lambda_(1) +lambda_(2) =3`
Also as a is perpendicular to c , therefore a, c=0
`rArr(2hat(i) + lambda_(1) hat(j) +3hatk)) . (vec(3hat(i) +6hat(j) + (lambda_(3) -1)hat(k)) =0`
`rArr 6 + 6lambda_(1) +3(lambda_(3) -1) =0`
`rArr 6lambda_(1) +3lambda_(3) + 3 =0`
`rArr 2lambda_(1) +lambda_(3) =-1`
Now from Eq . (i) `lambda_(2) =3 -2lambda` and from Eq . (ii)
`lambda_(3) = 2lambda_(1) -1`
`:. (lambda_(1) , lambda_(2) , lambda_(3)) -= (lambda_(1) , 3- 2lambda_(1) -1)`
If `lambda_(1) =- (1)/(2) ,` then
`lambda_(2) = 4 " and " lambda_(3) =0`
Thus a possible value of `(lambda_(1),lambda_(2) , lambda_(3)) = (-(1)/(2) , 4,0)`