Question

Find an equation of the plane. The plane through the point (3,0,1) and perpendicular to the line x=4t,y=4−t,z=5+6t

Answer 1

To find an equation of the plane through the point (3,0,1) and perpendicular to the line x=4t, y=4−t, z=5+6t, we need to find the normal vector of the plane by taking the cross product of the direction vector of the line and the normal vector. Then we can write the equation of the plane using the coefficients obtained.

Step-by-step explanation:

To find an equation of the plane through the point (3,0,1) and perpendicular to the line x=4t, y=4−t, z=5+6t, we need to find the normal vector of the plane. The direction vector of the line is ( 4, -1, 6), so the normal vector of the plane is perpendicular to this direction vector. We can take the cross product of the normal vector with the direction vector, which will give us the coefficients of the equation of the plane.

Let's find the cross product:

(4, -1, 6) x (x, y, z) = (0, 0, 0)

Using the cross product formula, we get -1z + 6y - 4x = 0. This equation represents the plane through the point (3,0,1) and perpendicular to the given line.