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
Find the parametric equations of the line through the origin that is orthogonal to the plane 2x+4y-z=0
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
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 = xā + at
y = yā + bt
z = zā + ct
where (xā, yā, zā) is a point on the line and (a, b, c) is the direction vector.
Since the line passes through the origin, its point (xā, yā, zā) 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.