1.

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

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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