Find the position vector of the foot of the perpendicular and the perpendicular distance from the point P with position vector
to the plane
Also, find the image of P in the plane.
Let the position vector of P be
so that
and M be the image of P in the plane
.
In addition, let Q be the foot of the perpendicular from P on to the given plane so that Q is the midpoint of PM.
Direction ratios of PM are proportional to 2, 1, 3 as PM is normal to the plane and parallel to
.
Recall the vector equation of the line passing through the point with position vector
and parallel to vector
is given by
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Here,
and![]()
Hence, the equation of PM is
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Let the position vector of M be
. As M is a point on this line, for some scalar α, we have
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Now, let us find the position vector of Q, the midpoint of PM.
Let this be
.
Using the midpoint formula, we have
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This point lies on the given plane, which means this point satisfies the plane equation
.
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We have the image ![]()
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Foot of the perpendicular![]()
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Using the distance formula, we have




Thus, the position vector of the image of the given point is
and that of the foot of perpendicular is
. Also, the length of this perpendicular is
units.