Let →a=^i+^j & →b=2^i+^j The point of intersection of the lines →r×→a=→b×→a&→r×→b=→a×→b is
A
−^i+^j+^k
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
−3^i−^j+^k
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3^i+^j−^k
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
^i−^j−^k
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B3^i+^j−^k →a=^i+^j & ^b=^2i−^k→r×→a=→b×→a⇒(→r−→b)×→a=0⇒→r=→b+λ→a
Similarly →r×→b=→a×→b⇒→r=→a+μ→b Substitute the vector →a & →b in (i) & (ii) and equating we get 2^i−^k+λ(^i+^j)=^i+^j+μ(2^i−^k)⇒2+λ=1+2μ,λ=1,μ=1∴ Point of intersection is 3^i+^j−^k