Q14 of 230 Page 30

Find the length and the foot of the perpendicular from the point (1, 3/2, 2) to the plane 2x – 2y + 4z + 5 = 0.

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





Let.


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



We have





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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