A plane P = 0 passing through the point 1,1 , 1 is perpendicular to the planes 2 x - y + 2 z = 5 and 3 x + 6 y - 2 z = 7 . If the distance of the point 1,2 , 3 from the plane P = 0 is k units, then the value of 34 k 2 is equal to

Question

A plane P = 0 passing through the point 1,1 , 1 is perpendicular to the planes 2 x - y + 2 z = 5 and 3 x + 6 y - 2 z = 7 . If the distance of the point 1,2 , 3 from the plane P = 0 is k units, then the value of 34 k 2 is equal to, 8 17, 16, 64, 128

MARKS App · Accepted Answer

Normal vector to the plane P = 0 is i ^ j ^ k ^ 2 - 1 2 3 6 - 2 = i ^ - 10 - j ^ - 10 + k ^ 15 = - 10 i ^ + 10 j ^ + 15 k ^ Equation of the plane through 1,1 , 1 is - 2 x - 1 + 2 y - 1 + 3 z - 1 = 0 ⇒ - 2 x + 2 y + 3 z - 3 = 0 So, the distance from the point 1,2 , 3 is - 2 + 4 + 9 - 3 17 = 8 17 units

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