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.
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.
To find a scalar equation for the plane that satisfies the given conditions, we need to find a normal vector for the plane.
First, let's find the direction vector of the line of intersection of the two planes. The direction vector can be obtained by taking the cross product of the normal vectors of the planes.
The normal vectors of the planes are:
Plane 1: n1 = <1, 2, 1>
Plane 2: n2 = <2, 1, 3>
To find the direction vector, we take the cross product of n1 and n2:
Direction vector = n1 x n2 = <(2*1) - (1*2), (1*1) - (3*2), (1*1) - (2*1)> = <-2, -5, -1>
Now that we have the direction vector of the line of intersection, we know that the normal vector of the plane we are looking for must be perpendicular to this vector. This means that the dot product between the normal vector of the plane and the direction vector of the line of intersection should be zero.
So, let the normal vector of the plane be <a, b, c>. Then we have the following equation:
<a, b, c> • <-2, -5, -1> = 0
Using the dot product formula, we can expand this equation:
(-2a) + (-5b) + (-c) = 0
Simplifying, we get:
-2a - 5b - c = 0
Finally, plugging in the coordinates of the point (1, 1, 4), we can solve for the scalar equation of the plane:
-2(1) - 5(1) - (4) = -2 - 5 - 4 = -11
Therefore, the scalar equation for the plane that passes through the point (1, 1, 4) and is perpendicular to the line of intersection of the given planes is:
-2x - 5y - z = 11
To determine a scalar equation for the plane, we first need to find the direction vector of the line of intersection of the two planes.
The direction vector of the line of intersection can be found by taking the cross product of the normal vectors of the two planes.
Given the equations of the planes:
Plane 1: x + 2y + z = 1
Plane 2: 2x + y + 3z = 3
Rewriting these equations in normal form:
Plane 1: N1 • P = 1
where N1 = <1, 2, 1> is the normal vector and P = <x, y, z>
Plane 2: N2 • P = 3
where N2 = <2, 1, 3> is the normal vector and P = <x, y, z>
We can now find the direction vector D by taking the cross product of N1 and N2:
D = N1 x N2
Calculating the cross product:
D = <1, 2, 1> x <2, 1, 3>
= <(2*1) - (1*3), (1*2) - (1*2), (1*1) - (2*1)>
= <-1, 0, -1>
Now that we have the direction vector of the line of intersection, we can write the equation of the plane using the point-normal form:
Equation: N • (P - P0) = 0
where N is the direction vector, P0 is the given point, and P = <x, y, z>
Let's substitute the given point (1, 1, 4) and the direction vector <-1, 0, -1> into the equation:
<-1, 0, -1> • (<x, y, z> - <1, 1, 4>) = 0
<-1, 0, -1> • <x-1, y-1, z-4> = 0
-1(x-1) + 0(y-1) - 1(z-4) = 0
-x + 1 - (z-4) = 0
-x - z + 3 = 0
Therefore, the 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 is -x - z + 3 = 0.