The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, -4, 3). Find the equation of the plane.
Given: the coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, -4, 3)
To find: the equation of the plane
As it is given that the foot of the perpendicular drawn from origin O to the plane is P(12, -4, 3)
This means that the required plane is passing through P(12, -4, 3) and is perpendicular to OP. Let the position vector of this point P be
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And it is also given the plane is normal to the line joining the points O(0,0,0) and P(12, -4, 3).
Then ![]()
Position vector of
- position vector of ![]()
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We know that the vector equation of a plane passing through the point
and perpendicular/normal to the vector
is given by
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Substituting the values from eqn(i) and eqn(ii) in the above equation, we get
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(by multiplying the two vectors using the formula
)
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is the vector equation of a required plane.
Let ![]()
Then, the above vector equation of the plane becomes,
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Now multiplying the two vectors using the formula
, we get
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This is the Cartesian form of the equation of the required plane.
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