Let →a=→i+→j+√1^k,→b=b1^i+b1^j+√2^k and →c=5^i+^j+√2^k be three vectors such that the projection vector of →b on →a is →a. If →a+→b is perpendicular to →c, the |→b| is equal to :
A
√22
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
√32
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D6 Projection of →b on →a=→a.→b|→b|=|→a|⇒b1+b2=2.......(1)and (→a+→b)⊥→c⇒(→a+→b.→c)=0⇒5b1+b2=−10......(2)from (1) and (2)⇒b1=−3 and b2=5 then |→b|=√b21+b22+2=6