1.

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

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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