If O be the origin and the coordinates of P be (1, 2, -3),then find the equation of the plane passing through P and perpendicular to OP.
If O be the origin and the coordinates of P be (1, 2, -3),then find the equation of the plane passing through P and perpendicular to OP.
Solution and Explanation
The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, -3) respectively.
Therefore, the direction ratios of OP are (1-0) =1, (2-0)=2, and (-3-0)=-3
It is known that the equation of the plane passing through the point(x1, y1, z1) is a(x-x1) + b(y-y1) + c(z-z1) = 0
where, a, b, and c are the direction ratios of normal.
Here, the direction ratios of normal are 1, 2, and -3, and the point P is (1, 2, -3).
Thus, the equation of the required plane is 1(x-1) + 2(y-2) - 3(z+3) = 0
⇒ x+2y-3z-14 = 0
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