Let the equation of the plane through the points (−1,1,1) and (1,−1,1) and perpendicular to the plane x+2y+2z=7 be kx+my−nz+p=0. Find k+m+n+p?
Open in App
Solution
Any plane through (−1,1,1) is A(x+1)+B(y−1)+C(z−1)=0 ...(1) It passes through (1,−1,1) ∴2A−2B+0C=0. ... (2) It is perpendicular to the plane x+2y+2z=7∴A+2B+2C=0. ... (3) Solving (2) and (3), we get A−4=B−4=C6 or A2=B2=C−3=a (say)
Thus, A,B and C are proportional to 2,2 and −3 respectively Substituting in (1), the equation of the plane is 2a(x+1)+2a(y−1)−3a(z−1)=0 ⇒2x+2y−3z+3=0 ⇒k+m+n+p=10