The equation of the plane passing through the lines x−41=y−31=z−22 and x−31=y−2−4=z5 is-
A
11x−y−3z=35
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
11x+y−3z=35
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
11x−y+3z=35
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D None of these. Normal vector of the plane is given by, →n=∣∣
∣
∣∣^i^j^k1121−45∣∣
∣
∣∣=9^i−3^j−5^k Hence equation of plane is, (→r−(3^i+2^j))⋅→n=0 ⇒9x−3y−5z−21=0