Let ∣→a∣=1,∣→b∣=2 and ∣→c∣=3 and →a,→b,→c be threee non-coplanar vectors. If ∣→d∣=4, then ∣[→d→b→c]→a+[→d→c→a]→b+[→d→a→b]→c[→a→b→c]∣ is equal to
A
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C2 [→a+→b→b+→c]⇒{(a+b)×→b×→c}⋅→c+→a⇒{→a×→b+→a×→c+→b×→b+→b×→c}⋅(→c+→a)⇒→a×→b⋅→c+→a×→c⋅→c+→b×→b⋅→c+→b×→c+→a×→b⋅→a+→a×→c⋅→a+→b×→b⋅+→b×→c⋅→a⇒[abc]+0+0+0+0+0+[bca]=2⇒[abc]Ans.