A test Question:
Find the parametric equations of the line through the origin that is orthogonal to the plane 2x+4y-z=0
the direction of a perpendicular to the given plane must be (2,4,-1)
so the vector equation must be
vector r = k (2,4,-1) + (0,0,0)
so in parametric form
x = 2k
y = 4k
z = -k
To find the parametric equations of the line through the origin that is orthogonal to the plane 2x + 4y - z = 0, we can use the normal vector of the plane.
The normal vector of the plane is given by the coefficients of x, y, and z in the equation of the plane. In this case, the normal vector is <2, 4, -1>.
A line that is orthogonal to the plane must have a direction vector that is perpendicular to the normal vector of the plane. So, let's find a direction vector for the line.
We can choose any vector that is perpendicular to the normal vector. One way to find a perpendicular vector is by taking the cross product of the normal vector with another vector that is not parallel to it.
Let's choose the vector <1, 0, 0> as a vector that is not parallel to the normal vector. Taking the cross product:
<1, 0, 0> x <2, 4, -1> = <-4, 1, 0>
So, the vector <-4, 1, 0> is perpendicular to the normal vector. This vector can be used as the direction vector for the line.
Now, we have a point on the line (the origin) and a direction vector. We can write the parametric equations of the line using the vector equation of a line in parametric form:
x = x₀ + at
y = y₀ + bt
z = z₀ + ct
where (x₀, y₀, z₀) is a point on the line and (a, b, c) is the direction vector.
Since the line passes through the origin, its point (x₀, y₀, z₀) is (0, 0, 0). And the direction vector is <-4, 1, 0>.
So, the parametric equations of the line are:
x = 0 + (-4)t = -4t
y = 0 + (1)t = t
z = 0 + (0)t = 0
Therefore, the parametric equations of the line through the origin that is orthogonal to the plane 2x + 4y - z = 0 are x = -4t, y = t, and z = 0.