The equation of the line passing through (−4,3,1), parallel to the plane x+2y−z−5=0 and intersecting the line x+1−3=y−32=z−2−1 is:
A
x+4−1=y−31=z−11
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x+41=y−31=z−13
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x+43=y−3−1=z−11
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
x+42=y−31=z−14
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Cx+43=y−3−1=z−11 Vector passing through (−4,3,1) and (−1,3,2)=3^i+^k
Normal vector of plane containing two intersecting lines is parallel to the vector. →r1=∣∣
∣
∣∣^i^j^k301−32−1∣∣
∣
∣∣=−2^i+6^k
∴ Required line is parallel to the vector. →r2=∣∣
∣
∣∣^i^j^k12−1−206∣∣
∣
∣∣=3^i−^j+^k ∴ Required equation of line passing through (−4,3,1) is x+43=y−3−1=z−11