Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane ![]()
Let point P = (1, 1, 2) and Q be the foot of the perpendicular drawn from to P the plane
.
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Direction ratios of PQ are proportional to 1, –2, 4 as PQ 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
![]()
Here,
and![]()
Hence, the equation of PQ is
![]()
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Let the position vector of Q be
. As Q is a point on this line, for some scalar α, 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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⇒ (1 + α)(1) + (1 – 2α)(–2) + (2 + 4α)(4) = –5
⇒ 1 + α – 2 + 4α + 8 + 16α = –5
⇒ 21α + 7 = –5
⇒ 21α = –12
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Foot of the perpendicular![]()
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Thus,![]()
Using the distance formula, we have




Thus, the required foot of perpendicular is
and the length of the perpendicular is
units.
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