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Equation of a plane perpendicular to a given vector and passing through a given point is considered one of the most asked concept.
22 Questions around this concept.
The coordinates of the foot of the perpendicular from the point (1, −2, 1) on the plane containing the lines
and
is
A vector in the first octant is inclined to the
- axis at
, to the
-axis at 45 and to the
-axis at an acute angle. If a plane passing through the points
and
, is normal to
,then
If the equation of the plane containing the line and perpendicular to the plane
is
then
is equal to:
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If the equation of the plane that contains the point and is perpendicular to each of the planes
and
is
then
The plane, passing through the points (0, -1, 2) and (-1, 2, 1) and parallel to the line passing through (5,1,-7) and (1,-1,-1), also passes through the point:
Imagine a pair of orthogonal vectors that share an initial point. Visualize grabbing one of the vectors and twisting it. As you twist, the other vector spins around and sweeps out a plane. Here, we describe that concept mathematically.
Let be a vector and P (x0, y0, z0) be a point. Then the set of all point Q (x, y, z) such that
orthogonal to
forms a plane.
We say that is a normal vector, or perpendicular to the plane. Remember, the dot product of orthogonal vectors is zero. This fact generates the vector equation of a plane:
If position vector of point P is and position vector of point Q is
, then
This is the vector equation of the plane.
Cartesian form
Thus, the coefficients of x, y and z in the cartesian equation of a plane are the direction ratios of the normal to the plane.
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