No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
→A×→B|→A|⋅|→B|sinθ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
→A×→B|→A|⋅|→B|cosθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
→A⋅→B|→A|⋅|→B|sinθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B→A×→B|→A|⋅|→B|sinθ We know, |→A×→B|=|→A|⋅|→B|sinθ
A vector divided by it's own magnitude gives unit vector along itself. ∴→A×→B|→A|⋅|→B|sinθ will be unit vector along →A×→B.