To find the equation of a plane, we need a point on the plane and a normal vector to the plane.
First, let's find the line of intersection of the two planes.
To do this, we need to find the direction vector of the line of intersection. This can be done by taking the cross product of the normal vectors of the two planes:
Normal vector of plane 1:
a = <1, 2, 1>
Normal vector of plane 2:
b = <2, 1, 3>
Direction vector of the line of intersection:
c = a x b
= <2, -1, -3>
Next, let's find a point on the line of intersection. To do this, we can set one of the variables to a constant (e.g. t), and solve the resulting system of equations:
From plane 1:
x + 2y + z = 1
2x + 4y + 2z = 2
Subtracting twice the equation of plane 1 from the second equation:
-2y - 3z = 0
y = -(3/2)z
Setting z = t:
y = -(3/2)t
x = 1 - 2y - z
x = 1 - 2(-(3/2)t) - t
x = 1 + 3t
So, a point on the line of intersection is (1 + 3t, -(3/2)t, t).
Now, we have a point on the plane (1, 1, 4) and a normal vector to the plane (2, -1, -3).
A scalar equation for the plane can now be found by using the point-normal form of the equation of a plane:
2(x - 1) - (y - 1) - 3(z - 4) = 0
Simplifying, we get:
2x - y - 3z + 7 = 0
So, a scalar equation for the desired plane is 2x - y - 3z + 7 = 0.
Determine a scalar equation for the plane that passes through the point (1, 1, 4) and is perpendicular to the line of intersection of the planes
x + 2y + z = 1 and 2x + y + 3z = 3.
1 answer