Question

find the equation of a plane perpendicular to the plane 2x-3y+6z+8=0 and passing through the intersection of x+2y+3z-4=0 and 2x-y-z+5=0

Answer 1

➡HERE IS YOUR ANSWER⬇

Since the plane passes through the intersection of the planes x+2y+3z-4=0 and 2x-y-z+5=0, let us consider the equation of the required field be :

(x+2y+3z-4) +k(2x-y-z+5) =0

or, (1+2k)x +(2-k)y +(3-k)z +(-4+5k) =0 ...(i)

which is perpendicular to another plane 2x-3y+6z+8=0.

Then, the sum of the multiplications of normal vectors of the two planes =0

or, 2(1+2k) -3(2-k) +6(3-k) =0

or, 2 + 4k - 6 + 3k + 18 - 6k = 0

or, k = -14

Therefore, the required plane is obtained from (i) as

(1-28)x + (2+14)y + (3+14)z + (-5-70) = 0

or, -27x + 16y + 17z - 75 = 0

or, 27x - 16y - 17z + 75 = 0 (Ans)

⬆HOPE THIS HELPS YOU⬅