a plane passes through the point P(4,0,0) and Q (0,0,4) and is parallel to the y axis find the distance of plane from origin
Answers
Answer:
Step-by-step explanation:
Consider the equation of the plane in intercept form
which is x/a + y/b + z/c = 1, where
a is x-intercept,
b is y-intercept
c is z-intercept.
Plane passes through P(4,0,0)
=> a =4
Plane passes through Q(0,0,4)
=> c =4
Substituting these, Equation of plane becomes x/4 + y/b + z/4 = 1
Dr's of the normal to the plane will be proportional to (1/4, 1/b, 1/4)
Given that plane is parallel to y-axis,
dr's of the y-axis are given by (0,1,0)
dr's of y-axis will be perpendicular to that of normal to the given plane
=>(1/4,1/b,1/4).(0,1,0) = 0
=> 1/b =0
Thus, equation of the given plane is x + z = 4.....(Ans)
Distance of a point (x₀ , y₀ , z₀ ) from the plane ax + by + cz + d = 0 is given by
|ax₀ + by₀ + cz₀ + d|/√a²+b²+c²
Distance of the plane from origin(0,0,0)
is |0+0-4|/√1 + 1
=4/√2
=2√2.