Class 12

Math

3D Geometry

Three Dimensional Geometry

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.

Any plane through $A(1,−1,2)$ is given by

$a(x−1)+b(y+1)+c(z−2)=0$ ........(i)

Since it is perpendicular to each of the planes $2x+3y−2z=5$ and $x+2y−3z=8$, we have

$2a+3b−2c=0$ .........(ii)

$a+2b−3c=0$ ......(iii)

On solving (ii) and (iii) by cross multiplication, we have

$(−9+4)a =(−2+6)b =(4−3)c =λ⇒a=−5λ,b=4λ,c=λ$

Putting these value in (i), we get the required equation as

$−5λ(x−1)+4λ(y+1)+λ(z−2)=0$

$⇒5(x−1)−4(y+1)−(z−2)=0$

$⇒5x−4y−z−7=0$

Distance of the point $P(−2,5,5)$ from this plane is given by

$d=5_{2}+(−4)_{2}+(−1)_{2} ∣5×(−2)−4×5−5−7∣ =42 ∣−42∣ =42 42 =42 $ units.