Question

Find the equation of a plane passing through the point (2, - 1, 5), perpendicular to the plane x + 2y - 3z = 7 and parallel to the line \(\frac{x+5}3\) = \(\frac{y+1}{-1}\) = \(\frac{z-2}{1}\).

Answer 1

Given – A plane passes through (2, - 1, 5), perpendicular to the plane x + 2y - 3z = 7 and parallel to the line \(\frac{x+5}3\) = \(\frac{y+1}{-1}\) = \(\frac{z-2}{1}\)

 To find – The equation of the plane 

Let the equation of the required plane be ax + by + cz + d = 0……(a) 

The plane passes through (2, - 1, 5) 

So, 2a - b + 5c + d = 0…………………(i) 

The direction ratios of the normal of the plane is given by (a, b, c) 

Now, this plane is perpendicular to the plane x + 2y - 3z = 7 having direction ratios (1, 2, - 3) 

So, a + 2b - 3c = 0……………(ii) 

This plane is also parallel to the line having direction ratios (3, - 1, 1) 

So, the direction of the normal of the required plane is also at right angles to the given line. 

So, 3a - b + c = 0………………(iii) 

Solving equations (ii) and (iii),

Putting these values in equation (i) we get, 

2Х(- α) - (- 10α) + 5(- 7α) + d = 0 i.e. d = 27α 

Substituting all the values of a, b, c and d in equation (a) we get,

-ax - 10ay - 7az + 27a = 0

⇒ x + 10y + 7z + 27 = 0