How do you find the parametric equations for the line through the point P = (4, -4, 1) that is perpendicular to the plane 3x + 1y - 4z = 1?

Question

Elijah Abram · Accepted Answer

#x=4+3t, y=-4+t, z=1-4t, t in RR.# Since the reqd. line is #bot# to the plane #: 3x+y-4z=1#, the direction vector #vecl# of the line is #||# to the normal #vecn# of the plane, Here, #vecn=(3,1,-4)#, and, we take #vecl=vecn#. Now, the Parametric Eqn. of a line through a pt.#(x_1,y_1,z_1)# and having direction vector #vecl=(l_1,l_2,l_3)# is given by, #(x,y,z)=(x_1,y_1,z_1)+t(l_1,l_2,l_3), t in RR#. # :. (x,y,z)=(4,-4,1)+t(3,1,-4), t in RR,# or, #x=4+3t, y=-4+t, z=1-4t, t in RR.#

Charles Anctil · Answer

undefined To find the parametric equations for the line through point P = (4, -4, 1) perpendicular to the plane 3x + y - 4z = 1, follow these steps:

1. Find the normal vector to the plane by extracting the coefficients of x, y, and z from the plane equation: (3, 1, -4).

2. Let's call the direction vector of the line we seek D. Since the line is perpendicular to the plane, the direction vector D should be parallel to the normal vector of the plane. Therefore, D = (3, 1, -4).

3. Now, to find a point on the line, use the given point P = (4, -4, 1).

4. The parametric equations of the line can be written as:
x = 4 + 3t
y = -4 + t
z = 1 - 4t

These equations represent all the points (x, y, z) on the line, where t is a parameter that can take any real value.