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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.

A 2x + 3y – 4z – 16 = 0 Correct Answer Incorrect Answer
B 2x + 3y – 4z – 19 = 0 Correct Answer Incorrect Answer
C 2x + 3y – 4z – 29 = 0 Correct Answer Incorrect Answer
D 2x + 3y – 4z – 39 = 0 Correct Answer Incorrect Answer

Solution

The coordinates of the points, O and P, are (0, 0, 0) and (2, 3, −4) respectively. Therefore, the direction ratios of OP are (2 − 0) = 2, (3 − 0) = 3, and (−4 − 0) = −4 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 2, 3, and −4 and the point P is (2, 3, −4). Thus, the equation of the required plane is 2(x – 2) + 3(y – 3) – 4(z + 4) = 0 => 2x + 3y – 4z – 29 = 0

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