Let point P = (1, 3/2, 2) and Q be the foot of the perpendicular drawn from P to the plane 2x – 2y + 4z + 5 = 0.
Direction ratios of PQ are proportional to 2, –2, 4 as PQ is normal to the plane.
Recall the equation of the line passing through (x1, y1, z1) and having direction ratios proportional to l, m, n is given by
Here, (x1, y1, z1) = (1,, 2) and (l, m, n) = (2, –2, 4)
Hence, the equation of PQ is
This point lies on the given plane, which means this point satisfies the plane equation.
⇒ 4α + 2 – (–4α + 3) + 16α + 8 + 5 = 0
⇒ 20α + 4α – 3 + 15 = 0
⇒ 24α = –12
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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