Q. 26

Mark against the A. 5x – 6y + 7z = 20

B. 7x – 6y + 5z = 72

C. 20x – 18y + 14z = 11

D. 10x – 18y + 28z = 13

Answer :

Given: Point A(2, 3, 4) lies on a plane which is parallel to 5x – 6y + 7z = 3


To find: Equation of the plane


Formula Used: Equation of a plane is


a(x – x1) + b(y – y1) + c(z – z1) = 0


where a:b:c is the direction ratios of the normal to the plane


(x1, y1, z1) is a point on the plane.


Explanation:


Since the plane (say P1) is parallel to the plane 5x – 6y + 7z = 3 (say P2), the direction ratios of the normal to P1 is same as the direction ratios of the normal to P2.


i.e., direction ratios of P1 is 5 : -6 : 7


Let the equation of the required plane be


a(x – x1) + b(y – y1) + c(z – z1) = 0


Here a = 5, b = -6 and c = 7


Since (2, 3, 4) lies on the plane,


5(x - 2) – 6 (y - 3) + 7 (z - 4) = 0


5x – 6y + 7z – 10 + 18 – 28 = 0


5x – 6y + 7z = 20


The equation of the plane is 5x – 6y + 7z = 20

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