Let `vec(a)=2vec(j)-3vec(k),vec(b) = hat(j)+3hat(k) and vec(c)=3vec(i)+3hat(j)+hat(k)`. Let `hat(n)` be a unit vector such that `vec(a).hat(n)=vec(b).hat(n)=0`. What is the value of `vec(c).hat(n)` ?

Options

A. `1`
B. `sqrt(19)`
C. `3`
D. `-3`

Answer

D

Explanation

As given `vec(a)=2hat(j) - 3 hat(k) and vec(b)=hat(j)+3hat(k)`
and `vec(c)=-3hat(i)+3hat(j)+hat(k)`
Let `vec(n) = x hat(i) + y hat(j) +z hat(k)`
Since `vec(a) and vec(n)` are perpendicular to each other.
`vec(a).hat(n)=0rArr (2 hat(i) - 3 hat(k) ) . (x hat(i)+ y hat(j) +zhat(k))=0`
`rArr 2y-3z=0 ...(1)`
and `vec(b) . hat(n) = 0 `
`rArr (hat(j)+3 hat(k)). (x hat(i) + y hat(j)+ z hat(k))=0`
`rArr y + 3 z = 0` ...(2)
On solving Eqs. (1) and (2)
y=z=0
Since `hat(n)` is a unit vector,
`sqrt(x^(2)+y^(2)+z^(2))=1 rArr x=1 ` [since, y=z=0]
hence, `hat(n) = hat(i)`
This gives, `vec(c). hat(n) = (-3hat(i) + 3 hat(j)+hat(k)). (hat(i))=-3`