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Question
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
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-x_1)+b(y-y_1)+c(z - z_1) = 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(z-1) +2(y-2) - 3(z+3) = 0
`=> x + 2y - 3z -14 = 0`
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