1
Question
If ABC is an equilateral triangle of side a, then the value of →AB.→BC+→BC.→CA+→CA.→AB is equal to
If ABC is an equilateral triangle of side a, then the value of →AB.→BC+→BC.→CA+→CA.→AB is equal to
Open in App
Solution
The correct option is C −3a22
Angle between vectors is measured by joining the initial points ⇒cos(−−→AB,−−→BC)=120°cos(−−→BC,−−→CA)=120°cos(−−→CA,−−→AB)=120° (−−→AB.−−→BC=|−−→AB||−−→BC|cos(−−→AB,−−→BC))=a2×∣∣∣−12∣∣∣=−a22−−→BC.−−→CA=|−−→BC||−−→CA|cos(−−→BC,−−→CA)=a2×∣∣∣−12∣∣∣=−a22−−→CA.−−→AB=|−−→CA||−−→AB|cos(−−→CA,−−→AB)=a2×∣∣∣−12∣∣∣=−a22∴−−→AB.−−→BC+−−→BC.−−→CA+−−→CA.−−→AB=−3a22
Angle between vectors is measured by joining the initial points
⇒cos(−−→AB,−−→BC)=120°
cos(−−→BC,−−→CA)=120°
cos(−−→CA,−−→AB)=120° (−−→AB.−−→BC=|−−→AB||−−→BC|cos(−−→AB,−−→BC))
=a2×∣∣∣−12∣∣∣=−a22
−−→BC.−−→CA=|−−→BC||−−→CA|cos(−−→BC,−−→CA)
=a2×∣∣∣−12∣∣∣=−a22
−−→CA.−−→AB=|−−→CA||−−→AB|cos(−−→CA,−−→AB)
=a2×∣∣∣−12∣∣∣=−a22
∴−−→AB.−−→BC+−−→BC.−−→CA+−−→CA.−−→AB=−3a22
Suggest Corrections
0
Similar questions
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
