Question:
Options:
- Only I
- Neither I nor II
- Both I and II
- Only II
- Only I
- Neither I nor II
- Both I and II
- Only II
Correct Answer: c)
Explanation:
‘ Gi√en, @ +5+2=0..(c) ‘
‘ I. Consider abé = @ – (b x €) ‘
! =—(b+8)- (bx &) …using Eq.(i) !
=—(b- (bx) +¢-(bxd))
77; 7
! = (be) + be)) !
=0+0)=0
1 Thus, the √ectors are coplanar. 1
hon a
‘ II. Consider, ‘
axb=ax(—a—2)
‘ = —(a x a+ ©) using Eq. (i)
‘ =—(@xd+axd) ‘
1 =—(0+axd)=—-(ax?o
‘ =¢ x a..(ii)
‘ using Eq Similarly, ‘
bxé=axb
‘ From Egs. (ii) and (iii), we get
‘ ixb=bxé=¢cxa
‘ Gi√en, @ +5+2=0..(c) ‘
‘ I. Consider abé = @ – (b x €) ‘
! =—(b+8)- (bx &) …using Eq.(i) !
=—(b- (bx) +¢-(bxd))
77; 7
! = (be) + be)) !
=0+0)=0
1 Thus, the √ectors are coplanar. 1
hon a
‘ II. Consider, ‘
axb=ax(—a—2)
‘ = —(a x a+ ©) using Eq. (i)
‘ =—(@xd+axd) ‘
1 =—(0+axd)=—-(ax?o
‘ =¢ x a..(ii)
‘ using Eq Similarly, ‘
bxé=axb
‘ From Egs. (ii) and (iii), we get
‘ ixb=bxé=¢cxa