Determine a vector equation of the line of intersection of the given planes.
x+3y-4z=-1
x+4z=1
First, we need to find the direction vector of the line by taking the cross product of the normal vectors of the planes.
The normal vector of the first plane is `<1, 3, -4>` and the normal vector of the second plane is `<1, 0, 4>`.
Taking the cross product:
`<1, 3, -4> X <1, 0, 4> = <-12, -4, -1>`
This is the direction vector of the line of intersection.
Next, we need to find a point on the line. We can do this by setting one of the variables to a constant and solving for the other two. For example, if we set z = 0, we can solve for x and y in the first plane:
x + 3y = -1
Choosing x = 0, we get:
3y = -1
y = -1/3
So a point on the line is (0, -1/3, 0).
Putting this together, the vector equation of the line is:
`<x, y, z> = <0, -1/3, 0> + t<-12, -4, -1>`
where t is a parameter.
To find the vector equation of the line of intersection between the two planes, we need to find a point on the line and a direction vector for the line.
First, let's find a point on the line. We can do this by setting one variable (in this case, z) to zero and solving for the other variables. Let's set z = 0:
From the first plane: x + 3y - 4z = -1
x + 3y = -1
x = -1 - 3y
Substituting z = 0, we get x = -1 - 3y.
From the second plane: x + 4z = 1
x + 4(0) = 1
x = 1
Therefore, a point on the line of intersection is (1, 0, 0).
Next, let's find the direction vector for the line. To do this, we can take the cross product of the normal vectors of the two planes.
The normal vector of the first plane is (1, 3, -4).
The normal vector of the second plane is (1, 0, 4).
Taking the cross product of these two vectors, we get the direction vector of the line:
(1, 3, -4) x (1, 0, 4) = [(3)(4) - (-4)(0), (-4)(1) - (1)(4), (1)(0) - (3)(1)]
= (12, -8, -3)
Now, we can write the vector equation of the line of intersection:
r = (1, 0, 0) + t(12, -8, -3)
where r is the position vector of any point on the line, and t is a scalar parameter.