# Write the equation of the plane passing through point A perpendicular to the vector

Write the equation of the plane passing through point A perpendicular to the vector ВС А (0; -8; 10) В (-5; 5; 7) С (-8; 0; 4)

Given: A (0; -8; 10), B (-5; 5; 7), C (-8; 0; 4).

It is necessary: ​​Write – the equation of the plane passing through a given point A, perpendicular to the vector BC.

Answer: BC will be a normal vector (that is, a vector perpendicular to the plane), the equation of a plane passing through a given point, perpendicular to the normal, has the form:

a (x – x0) + b (y – y0) + c (z – z0) = 0 where a, b, c are the coordinates of the BC vector (in our case it is (3; 5; 3)), and x0, y0, z0 is the coordinate of the point through which the plane goes, in our case it is point A.

Substitute and get: 3 (x – 0) + 5 (y +8) + 3 (z – 10) = 0, open the brackets and get: 3x + 5y + 3z + 10 = 0.

When calculating the coordinates of the vector BC, the coordinates of the beginning are subtracted from the coordinates of the end, we have BC: (-5 + 8; 5 – 0; 7 – 4), i.e. (3; 4; 3).

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