Find the equation of the plane that contains the point A(1,−1,2) and is perpendicular to both the planes 2x+3y−2z=5 and x+2y−3z=8. Hence, find the distance of the point P(−2,5,5) from the plane obtained above.

Question

Filo · Accepted Answer

Any plane through A(1,−1,2) is given bya(x−1)+b(y+1)+c(z−2)=0 ........(i)Since it is perpendicular to each of the planes 2x+3y−2z=5 and x+2y−3z=8, we have2a+3b−2c=0 .........(ii)a+2b−3c=0 ......(iii)On solving (ii) and (iii) by cross multiplication, we have(−9+4)a​=(−2+6)b​=(4−3)c​=λ⇒a=−5λ,b=4λ,c=λPutting these value in (i), we get the required equation as−5λ(x−1)+4λ(y+1)+λ(z−2)=0⇒5(x−1)−4(y+1)−(z−2)=0⇒5x−4y−z−7=0Distance of the point P(−2,5,5) from this plane is given byd=52+(−4)2+(−1)2​∣5×(−2)−4×5−5−7∣​=42​∣−42∣​=42​42​=42​ units.

Question Text	Find the equation of the plane that contains the point $A (1, - 1, 2)$ and is perpendicular to both the planes $2 x + 3 y - 2 z = 5$ and $x + 2 y - 3 z = 8$ . Hence, find the distance of the point $P (- 2, 5, 5)$ from the plane obtained above.
Answer Type	Text solution:1
Upvotes	150

Find the equation of the plane that contains the point $A (1, - 1, 2)$ and is perpendicular to both the planes $2 x + 3 y - 2 z = 5$ and $x + 2 y - 3 z = 8$ . Hence, find the distance of the point $P (- 2, 5, 5)$ from the plane obtained above.

Text solutionVerified

Practice questions from similar books