Question

Find parametric equations of the line passing through point P(-4, 1, 5) that is perpendicular to the plane of equation 4x – 5y + z = 7. Consider points P, Q, and R. P(-2, 1, 5), Q(3, 1, 3), and R(-2, 1, 0) . Find the general equation of the plane passing through P, Q, and R.

Answer 1

To find the parametric equations of the line passing through a point that is perpendicular to a given plane, find the normal vector of the plane and take the cross product of the normal vector with another vector on the line.

Step-by-step explanation:

To find the parametric equations of the line passing through point P(-4, 1, 5) that is perpendicular to the plane of equation 4x – 5y + z = 7, we need to first find the normal vector of the plane.

The coefficients of x, y, and z in the equation represent the components of the normal vector. So the normal vector is N(4, -5, 1).

Next, we can find the direction vector of the line by taking the cross product of the normal vector and another vector. Let's take the cross product of N(4, -5, 1) and the vector connecting two points on the line, such as P(1, 1, 5) and Q(2, -2, 3). The cross product gives us the direction vector D(-10, 8, -13).

Finally, we can write the parametric equations of the line as:

x = -4 + (-10)t

y = 1 + 8t

z = 5 + (-13)t

Answer 2

To find the parametric equations of the line passing through point P(-4, 1, 5) that is perpendicular to the plane of equation 4x – 5y + z = 7, we need to find the vector that is perpendicular to the plane. The vector perpendicular to the plane is n = (4, -5, 1). The parametric equations of the line passing through P are x = -4 + 4t, y = 1 - 5t, z = 5 + t.

Step-by-step explanation:

To find the parametric equations of the line passing through point P(-4, 1, 5) that is perpendicular to the plane of equation 4x – 5y + z = 7, we need to find the vector that is perpendicular to the plane.

The coefficients of x, y, and z in the plane equation give the components of the normal vector to the plane. So the vector perpendicular to the plane is n = (4, -5, 1).

We can use the parametric form of a line to find the equations. Let t be a parameter. The line passing through P can then be expressed as: x = -4 + 4t, y = 1 - 5t, z = 5 + t.