Question

-Find the equation of the plane Containing the
line 9 + 7 = 1,20 and Parallel to the
line at Z=1,420​

Answer 1

Answer:The required line is the intersection of planes

2x−y+z−3=0    ...(1)

and 3x+y+z−5=0     ...(2)

Thus the plane containing required line is  

(2x−y+z−3)+λ(3x+y+z−5)=0

i.e., (3λ+2)x+(λ−1)y+(λ+1)z+(−5λ−3)=0

As its distance from point (2,1,−1) is  

6

​  

 

1

​  

 

⇒  

∣

∣

∣

∣

∣

​  

 

11λ  

2

+12λ+6

​  

 

(3λ+2)2+(λ−1)1−(λ+1)+(−5λ−3)

​  

 

∣

∣

∣

∣

∣

​  

=  

6

​  

 

1

​  

 

⇒λ(5λ+24)=0

⇒λ=0,−  

5

24

​  

 

∴ Required planes are 2x−y+z−3=0 and  

62x+29y+19z−105=0

How satisfied are yo

Step-by-step explanation: