The distance of the point (1, -2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x - y + 2z = 3 and 2x - 2y + z + 12 = 0, is :

Question

The distance of the point (1, -2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x - y + 2z = 3 and 2x - 2y + z + 12 = 0, is : (A) 2 √2 (B) 2 (C) √2 (D) (1/√2)

Tardigrade Admin · Accepted Answer

Let equation of plane be a(x-1)+b(y-2)+c(z-2)=0(1) i) is perpendicular to given planes then a-b+2c=0 2a-2b+c=0 Solving above equations c = 0 and a = b equation of plane 1) can be x + y - 3 = 0 distance from (1,-2,4) will be D=(|1-2-3|/√1+1)=(4/√2)=2√2

Q. The distance of the point $(1, - 2, 4)$ from the plane passing through the point $(1, 2, 2)$ and perpendicular to the planes $x - y + 2 z = 3$ and $2 x - 2 y + z + 12 = 0$ , is :

$22$

2

$2$

$\frac{1}{2}$

Q. The distance of the point (1,−2,4) from the plane passing through the point (1,2,2) and perpendicular to the planes x−y+2z=3 and 2x−2y+z+12=0, is :

22​

2

2​

2​1​

Let equation of plane be a(x−1)+b(y−2)+c(z−2)=0_____(1) i) is perpendicular to given planes then a−b+2c=0 2a−2b+c=0 Solving above equations c=0 and a=b equation of plane 1) can be x + y - 3 = 0 distance from (1,-2,4) will be D=1+1​∣1−2−3∣​=2​4​=22​

Q. The distance of the point $(1, - 2, 4)$ from the plane passing through the point $(1, 2, 2)$ and perpendicular to the planes $x - y + 2 z = 3$ and $2 x - 2 y + z + 12 = 0$ , is :

$22$

$2$

$\frac{1}{2}$