Find the equation of the plane through the straight line x+y-2z +4 =0=3x-y+2z+1 and parallel to the straight line (x+2)/2= (y-2)/2= (z-1)/-1

Question

Deepak Kumar Shringi · Answer

hereby attaching same question with different data i hope this will help you

Equation of the planes are2x+y−z=3and5x−3y+4z+9=0
The equation of the plane passing through the line of intersection of these planes is
(2x+y−z−3)+λ(5x−3y+4z+9)=0
x(2+5λ)+y(1−3λ)+2(4λ−1)+9λ−3=0-----(1)
Step 2:
The plane is parallel to the line
x−12=y−34=z−55
∴2(2+5λ)+4(1−3λ)+5(4λ−1)=0
(i.e)18λ+3=0
⇒λ=−6
Step 3:
Put the value ofλλin emu(1) we obtain
x(2−56)+y(1+36)+z(−46−1)−96−3=0
⇒7x+9y−10z−27=0
This is the equation of the required plane.

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Find the equation of the plane through the straight line x+y-2z +4 =0=3x-y+2z+1 and parallel to the straight line (x+2)/2= (y-2)/2= (z-1)/-1

Find the equation of the plane through the straight line x+y-2z +4 =0=3x-y+2z+1 and parallel to the straight line (x+2)/2= (y-2)/2= (z-1)/-1

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