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.
Determine a vector equation of the line of intersection of the given planes.
x+3y-4z=-1
x+4z=1
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