If the distance of the point P (1,-2,1) from the plane x +2 y -2 z =α, where α0, is 5 , then the foot of the perpendicular from P to the plane is

Question

If the distance of the point P (1,-2,1) from the plane x +2 y -2 z =α, where α>0, is 5 , then the foot of the perpendicular from P to the plane is (A) ((8/3), (4/3),-(7/3)) (B) ((4/3),-(4/3), (1/3)) (C) ((1/3), (2/3), (10/3)) (D) ((2/3),-(1/3), (5/2))

Tardigrade Admin · Accepted Answer

Distance of point (1,-2,1) from plane x +2 y -2 z =α is 5 ⇒ α=10 Equation of PQ (x-1/1)=(y+2/2)=(z-1/-2)=t Q ≡( t +1,2 t -2,-2 t +1) and PQ =5 ⇒ t =(5+α/9)=(5/3) ⇒ Q ≡((8/3), (4/3), (-7/3))

Q. If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is $5$ , then the foot of the perpendicular from $P$ to the plane is

$(\frac{8}{3}, \frac{4}{3}, - \frac{7}{3})$

$(\frac{4}{3}, - \frac{4}{3}, \frac{1}{3})$

$(\frac{1}{3}, \frac{2}{3}, \frac{10}{3})$

$(\frac{2}{3}, - \frac{1}{3}, \frac{5}{2})$

Q. If the distance of the point P(1,−2,1) from the plane x+2y−2z=α, where α>0, is 5 , then the foot of the perpendicular from P to the plane is

(38​,34​,−37​)

(34​,−34​,31​)

(31​,32​,310​)

(32​,−31​,25​)

Distance of point (1,−2,1) from plane x+2y−2z=α is 5 ⇒α=10 Equation of PQ 1x−1​=2y+2​=−2z−1​=t Q≡(t+1,2t−2,−2t+1) and PQ=5 ⇒t=95+α​=35​ ⇒Q≡(38​,34​,3−7​)

Q. If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is $5$ , then the foot of the perpendicular from $P$ to the plane is

$(\frac{8}{3}, \frac{4}{3}, - \frac{7}{3})$

$(\frac{4}{3}, - \frac{4}{3}, \frac{1}{3})$

$(\frac{1}{3}, \frac{2}{3}, \frac{10}{3})$

$(\frac{2}{3}, - \frac{1}{3}, \frac{5}{2})$