If →a=3^j−4^k →b=2^i−^j+2^k and →c is directed along the direction parallel to angle bisector of →a and →b and the projection of →c on →d is of magnitude 109 where →d=4^i−4^j+7^k then |→c|2 is
A
81
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
120
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
108
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
150
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B120 direction of angle bisector of →a and →b can be calculated as →a|→a|+→b|→b| (because resultant of unit vector will lie in direction parallel to angle bisector) =3^j−4^k√(3)2+(−4)2+2^i−^j+2^k√(2)2+(−1)2+(2)2 =3^j−4^k5+2^i−^j+2^k3 =10^i+4^j−2^k15 →c=λ×10^i+4^j−2^k15 and also →c.→d|→d|=109 ⇒(λ×10^i+4^j−2^k15).(4^i−4^j+7^k√(4)2+(−4)2+(7)2)=109 ⇒λ×(40−16−14)15×9=109 ⇒λ=15 →c=15×10^i+4^j−2^k15 →c=10^i+4^j−2^k |→c|2=120