Q. 13

# Mark against the

Given: Plane passes through A(2, 2, 1) and B(9, 3, 6). Plane is perpendicular to 2x + 6y + 6z = 1

To find: Equation of the plane

Formula Used: Equation of a plane is

a(x – x1) + b(y – y1) + c(z – z1) = 0

where a:b:c is the direction ratios of the normal to the plane.

(x1, y1, z1) is a point on the plane.

Explanation:

Let the equation of plane be a(x – x1) + b(y – y1) + c(z – z1) = 0

Since (2, 2, 1) is a point in the plane,

a(x – 2) + b(y – 2) + c(z – 1) = 0 … (1)

Since B(9, 3, 6) is another point on the plane,

a(9 – 2) + b(3 – 2) + c(6 – 1) = 0

7a + b + 5c = 0 … (1)

Since this plane is perpendicular to the plane 2x + 6y + 6z = 1, the direction ratios of the normal to the plane will also be perpendicular.

So, 2a + 6b + 6c = 0 a + 3b + 3c = 0 … (2)

Solving (1) and (2),   a : b : c = 3 : 4 : -5

Substituting in (1),

3x – 6 + 4y – 8 – 5z + 5 = 0

3x + 4y – 5z – 9 = 0

Therefore the equation of the plane is 3x + 4y – 5z – 9 = 0

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