Determine an equation of the line of intersection of the planes 4x − 3y − z = 1 and 2x + 4y + z = 5.
Add the 2 equations ...
6x + y = 6
y = 6-6x
let x = 0 , y = 6
2(0) + 4(6) + z = 5 --> z = -19)
we have a point (0,6,-19) on the line of intersection
let x = 1, y = 0
2(1) + 4(0) + z = 5 --> z = 3
and (1,0,3) is another point on that line
So a direction vector of that line is (1, -6, 22)
and using the point (1,0,3)
x = 1 + t
y = -6t
z = 3 + 22t
or, in symmetric form:
(x-1)/1 = y/-6 = (z-3)/22
To determine the equation of the line of intersection of the two planes, we can solve the system of equations formed by setting the two plane equations equal to each other.
Let's solve the system step-by-step:
Step 1: Write down the two plane equations:
4x - 3y - z = 1
2x + 4y + z = 5
Step 2: Choose a variable to eliminate by adding or subtracting the two equations. In this case, we'll eliminate the variable z by adding the two equations together:
(4x - 3y - z) + (2x + 4y + z) = 1 + 5
Simplifying this equation gives:
6x + y = 6
Step 3: Solve the resulting equation for one variable. In this case, we'll solve for y:
y = 6 - 6x
Step 4: Rewrite the equation in vector form using the variables x and y:
x = x
y = 6 - 6x
Step 5: Combine the two equations into a single vector equation:
r = (x, 6 - 6x)
So, the equation of the line of intersection of the two planes is:
r = (x, 6 - 6x)
Alternatively, you can write it as:
x = t
y = 6 - 6t
z = 6t - 1
Where t is a parameter that can take any real value.
To find the equation of the line of intersection of the planes, we need to find two points that lie on the line. We can do this by setting one of the variables (say, z) to 0 and solving for the remaining variables (x and y).
Let's set z = 0 in both equations:
For the first plane, 4x - 3y - z = 1 becomes 4x - 3y - 0 = 1, which simplifies to 4x - 3y = 1. (Equation 1)
For the second plane, 2x + 4y + z = 5 becomes 2x + 4y + 0 = 5, which simplifies to 2x + 4y = 5. (Equation 2)
Now we have a system of linear equations (Equation 1 and Equation 2). We can solve this system to find the values of x and y.
Multiplying Equation 1 by 2, we get: 8x - 6y = 2. (Equation 3)
Subtracting Equation 2 from Equation 3, we eliminate x:
(8x - 6y) - (2x + 4y) = 2 - 5,
6x - 10y = -3. (Equation 4)
Now we have a new equation (Equation 4) with only two variables, x and y.
To find the values of x and y, we need another equation. Let's set y = 0 in Equation 4:
6x - 10(0) = -3,
6x = -3,
x = -3/6,
x = -1/2.
Now substitute the value of x into Equation 2:
2(-1/2) + 4y = 5,
-1 + 4y = 5,
4y = 5 + 1,
4y = 6,
y = 6/4,
y = 3/2.
So one point on the line of intersection is (-1/2, 3/2, 0).
Now, let's set x = 0 in Equation 4:
6(0) - 10y = -3,
-10y = -3,
y = -3/-10,
y = 3/10.
Substitute the value of y into Equation 1:
4x - 3(3/10) = 1,
4x - 9/10 = 1,
4x = 1 + 9/10,
4x = 10/10 + 9/10,
4x = 19/10,
x = 19/10 ÷ 4,
x = 19/10 ÷ 4/1,
x = 19/10 × 1/4,
x = 19/40.
So another point on the line of intersection is (19/40, 3/10, 0).
Now, we have two points on the line of intersection: (-1/2, 3/2, 0) and (19/40, 3/10, 0).
To find the direction vector of the line, subtract the coordinates of the two points:
Direction vector = (-1/2 - 19/40, 3/2 - 3/10, 0 - 0),
Direction vector = (-20/40 - 19/40, 30/40 - 12/40, 0 - 0),
Direction vector = (-39/40, 18/40, 0).
The equation of the line of intersection can be written as:
x = -1/2 - 39/40t,
y = 3/2 + 18/40t,
z = 0.
So the equation of the line of intersection of the planes 4x − 3y − z = 1 and 2x + 4y + z = 5 is:
x = -1/2 - 39/40t,
y = 3/2 + 18/40t,
z = 0,
where t is a parameter that represents any real number.