Q. 11

# Mark against the correct answer in each of the following:The equation of a plane through the point A(1, 0, -1) and perpendicular to the line isA. 2x + 4y – 3z = 3B. 2x – 4y + 3z = 5C. 2x + 4y – 3z = 5D. x + 3y + 7z = -6

Given: Plane passes through the point A(1, 0, -1).

Plane is perpendicular to the line To find: Equation of the plane.

Formula Used: Equation of a plane is ax + by + cz = d where (a, b, c) are the direction ratios of the normal to the plane.

Explanation:

Let the equation of the plane be

ax + by + cz = d … (1)

Substituting point A,

a – z = d

Since the given line is perpendicular to the plane, it is the normal.

Direction ratios of line is 2, 4, -3

Therefore, 2 + 3 = d

d = 5

So the direction ratios of perpendicular to plane is 2, 4, -3 and d = 5

Substituting in (1),

2x + 4y – 3z = 5

Therefore, equation of plane is 2x + 4y – 3z = 5

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Find the coordinate of the point P where the line through and crosses the plane passing through three points and Also, find the ratio in which P divides the line segment AB.

Mathematics - Board Papers