Let the unit vector be r=xi+yj+zk and a=i+j+3k,b=i+3j+k and c=i+j+k Given, [
r
ab
]=0, i.e., coplanar. ⇒ |
x
y
z
1
1
3
1
3
1
|=0 ⇒ x(1−9)−y(1−3)+z(3−1)=0 ⇒ −8x+2y+2x=0 ⇒ −4x+y+z=0 .......(i) and r ⋅ c = 0, i.e., perpendicular ⇒ (xi+yj+zk).(i+j+k)=0 ⇒ x+y+z=0 .......(ii) On solving Eqs. (i) and (ii), we get 5y+5z=0 ⇒ y=−z ........(iii) ∵r is a unit vector. ∴ |r|=1=√x2+y2+z2 ⇒ x2+y2+z2=1 ⇒ x2+2v2=1 [from Eq. (iii)] …(iv) Put y = -z in Eq. (i), we get −4x=0⇒x=0 From Eq. (iv), we get 2y2=1⇒y=±