Question

find the equation of plane through the points (2,1,2)and (1,3,-2) and parallel to the x axis
​

Answer 1

We have to find the equation of plane through the point (2, 1, 2) and (1, 3, -2) and parallel to x -axis.

solution : let P = (2, 1, 2) and Q = (1, 3, -2)

so, PQ = (2, 1, 2) - (1, 3, -2) = (1, -2, 4)

vector direction of x - axis , u = (1, 0, 0)

now normal vector, n = u × PQ

= (1, 0 , 0) × (1, -2, 4)

= i × (i - 2j + 4k)

= -2k - 4j

so normal vector = (0, -4, -2)

now equation of plane :

l(x - a) + m(y - b) + n(z - c) = 0

⇒0(x - 2) -4(y - 1) - 2(z - 2) = 0

⇒4y - 4 + 2z - 4 = 0

⇒2y + z - 4 = 0

Therefore the equation of plane is 2y + z = 4