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If O is the origin and P(1, 2, -3) is a given point, then the equation of the plane through P and perpendicular to OP is A. x + 2y – 3z = 14 B. x – 2y + 3z = 12 C. x – 2y – 3z = 14 D. None of these |
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Answer» Given: P(1, 2, -3) is a point on the plane. OP is perpendicular to the plane. Explanation: Let equation of plane be ax + by + cz = d … (1) Substituting point P, ⇒ a + 2b -3c = d … (2) \(\overset\rightarrow{OP}\) = \(\hat{i}\) + 2\(\hat{j}\) - 3\(\hat{k}\) Since OP is perpendicular to the plane, direction ratio of the normal is (1, 2, -3) Substituting in (2) 1 + 4 + 9 = d d = 14 Substituting the direction ratios and value of ‘d’ in (1), we get x + 2y – 3z = 14 Therefore equation of plane is x + 2y – 3z = 14 |
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