Find the distance of the point (1, –2, 4) from a plane passing through the point (1, 2, 2) and perpendicular to the planes x – y + 2z = 3 and 2x – 2y + z + 12 = 0.
We know, equation of plane passing through (x1, y1, z1):
a(x – x1) + b(y – y1) + c(z – z1) = 0
⟹ Equation of plane passing through (1, 2, 2):
a(x – 1) + b(y – 2) +c(z – 2) = 0
i.e. ax + by + cz = a +2b +2c eq(i)
We know, if two planes a1x + b1y + c1z + d1 = 0 and
a2x + b2y + c2z + d2 = 0 are perpendicular, then:
a1.a2 + b1.b2 + c1.c2 = 0
According to question,
⟹ (1)(a) + (–1)(b) + (2)(c) = 0
⟹ (2)(a) + (–2)(b) + (1)(c) = 0
i.e.
⟹ a – b + 2c = 0
⟹ 2a – 2b + c = 0
Solving the above equations using cross multiplication method:
⟹
⟹ a = 3θ , b = 3θ , c = 0
Putting this in eq(i)
Equation of plane:
3θ(x) + 3θ(y) + (0)z = 3θ + 2(3θ) + 0
i.e.
x + y = 3
Distance of (1, –2, 4) from x + y =3
We know, the distance of point (x1,y1,z1) from the plane
is given by:
Putting the necessary values,
⟹
⟹
∴ the distance of the point (1, –2, 4) from plane passing through the point (1, 2, 2) and perpendicular to the planes x – y + 2z = 3 and
2x – 2y + z + 12 = 0 is units.