Let the foot of perpendicular of the point P 3 , – 2 , – 9 on the plane passing through the points – 1 , – 2 , – 3 , 9 , 3 , 4 , 9 , – 2 , 1 be Q α , β , γ . Then the distance Q from the origin is

Question

Let the foot of perpendicular of the point P 3 , – 2 , – 9 on the plane passing through the points – 1 , – 2 , – 3 , 9 , 3 , 4 , 9 , – 2 , 1 be Q α , β , γ . Then the distance Q from the origin is, 42, 38, 35, 29

MARKS App · Accepted Answer

Given that three points lie on a plane, A C → = 10 i ^ + 4 k ^ A B → = 10 i ^ + 5 j ^ + 7 k ^ A C → × A B → = i ^ j ^ k ^ 10 0 4 10 5 7 ⇒ A C → × A B → = - 20 i ^ - 30 j ^ + 50 k ^ ⇒ A C → × A B → = 2 i ^ + 3 j ^ - 5 k ^ Equation of plane will be, ⇒ 2 x + 3 y - 5 z = d And the plane passing through, – 1 , – 2 , – 3 ⇒ - 2 - 6 + 15 = d ⇒ d = 7 ∴ 2 x + 3 y - 5 z = 7 Now finding the foot of perpendicular we get, x - 3 2 = y + 2 3 = z + 9 - 5 = - 38 38 x = 1 , y = - 5 , z = - 4 Q 1 , – 5 , – 4 Distance from origin = 1 + 25 + 16 = 42 Hence this is the required option.

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