Question

Find the equation of plane through the point (2,3,5) and perpendicular to 2x-3y+z and 4x+y-3z=0

Answer 1

Answer:

 The equation of the required plane is

  4x + 5y + 7z = 58

Solution:

  Since the required plane passes  through the point (2, 3, 5), we consider the plane as

  A (x - 2) + B (y - 3) + C (z - 5) = 0 ... (1)

  Also given that (1) no. plane is perpendicular to the planes 2x - 3y + z = 0 and 4x + y - 3z = 0. Then

  2A - 3B + C = 0 ... (2)

  4A + B - 3C = 0 ... (3)

  Now, eliminating A, B, C from (1), (2) and (3) no. equations, we get

$\left|\begin{array}{ccc}x-2 & y-3 & z - 5\\ 2 & -3 & 1\\4 & 1 & -3\end{array}\right|=0$

or, (x - 2) (9 - 1) - (y - 3) (- 6 - 4) + (z - 5) (2 + 12) = 0

[ expanding along the first row ]

or, 8 (x - 2) - (- 10) (y - 3) + 14 (z - 5) = 0

or, 8x - 16 + 10y - 30 + 14z - 70 = 0

or, 8x + 10y + 14z = 116

or, 4x + 5y + 7z = 58

    Thus the equation of the required plane is

    4x + 5y + 7z = 58