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The equation of a plane through the point A(1, 0, -1) and perpendicular to the line \(\frac{x+1}2\) = \(\frac{y+3}4\) = \(\frac{z+7}{-3}\) isA. 2x + 4y – 3z = 3 B. 2x – 4y + 3z = 5 C. 2x + 4y – 3z = 5 D. x + 3y + 7z = -6 |
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Answer» Given: Plane passes through the point A(1, 0, -1). Plane is perpendicular to the line \(\frac{x+1}2\) = \(\frac{y+3}4\) = \(\frac{z+7}{-3}\) To find: Equation of the plane. Formula Used: Equation of a plane is ax + by + cz = d where (a, b, c) are the direction ratios of the normal to the plane. Explanation: Let the equation of the plane be ax + by + cz = d … (1) Substituting point A, a – z = d Since the given line is perpendicular to the plane, it is the normal. Direction ratios of line is 2, 4, -3 Therefore, 2 + 3 = d d = 5 So the direction ratios of perpendicular to plane is 2, 4, -3 and d = 5 Substituting in (1), 2x + 4y – 3z = 5 Therefore, equation of plane is 2x + 4y – 3z = 5 |
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