If |→a|=1,|→b|=2,|→c|=3 and →a+→b+→c=0, then the value of →a.→b+→b.→c+→c.→a equals
A
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
−7
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
−5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
−143
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B−7 |→a|=1,∣∣→b∣∣=2,|→c|=3→a+→b+→c=0∣∣→a+→b+→c∣∣2=0(→a+→b+→c)(→a+→b+→c)=0→a.→a+→a.→b+→a.→c+→b.→a+→b.→b+→b.→c+→c.→a+→c.→b+→c.→c=0|→a|2+∣∣→b∣∣2+|→c|2+2(→a.→b+→b.→c+→c.→a)=012+(2)2+32+2(→a.→b+→b.→c+→c.→a)=0→a.→b+→b.→c+→c.→a=−7