Question

A plane passes through (1,-2,1) and is perpendicular to the planes 2x - 2y + z = 0 and x - y + 2z = 4. The distance of the point (1, 2, 2) from this plane is______ units.

पर्याय

• 1

• √2

• 2√2

• √3

Answer 1

A plane passes through (1,-2,1) and is perpendicular to the planes 2x - 2y + z = 0 and x - y + 2z = 4. The distance of the point (1, 2, 2) from this plane is 2√2 units.

Explanations:

The equation of a plane passing through (1, − 2, 1) is
a(x − 1) + b(y + 2) + c(z − 1) = 0 ...(i)
Plane (i) is perpendicular to planes
2x − 2y + z = 0 and x − y + 2z = 4.
∴ 2a − 2b + c = 0, and ...(ii)
a − b + 2c = 0 ...(iii)
Solving (ii) and (iii), we get
a = −3, b = −3, c = 0
Substituting the values of a, b, and c in equation (i),
we get
x + y + 1 = 0
∴ The distance of this plane from (1, 2, 2) is

\[\mathbf{d}=\left|\frac{1+2+1}{\sqrt{1+1}}\right|=2\sqrt{2}\]