​ Find the vector equation of the plane passing through the intersection of the planes .(i + j + k) = 6 ​

Question

Find the vector equation of the plane passing through the intersection of the planes \(\bar r\).(i + j + k) = 6 and \(\bar r\).(2i + 3 j + 4k) = -5 at the point (1,1,1).

Answer 1

The Cartesian equation of the planes are x + y + z = 6 and 2x + 3y + 4z = – 5. Therefore the equation of the plane passing through the intersection of these planes is

x + y + z – 6 + k(2x + 3y + 4z + 5) = 0

Since it pass through (1, 1, 1) we get,

1 + 1 + 1 – 6 + k(2 + 3 + 4 + 5) = 0 ⇒ -3 + k14 ⇒ k = \(\frac{3}{14}\)

∴ the equation is

x + y + z + -6 + \(\frac{3}{14}\) (2x + 3 y + 4z + 5) = 0

14x + 14y + 14z – 84 + 6x + 9y + 12z + 15 = 0

20x + 23y + 26z = 69

Vector equation is \(\bar r\). (20i + 23j + 26k) = 69.