q4 Find parametric equations for the line through the point

q4

Find parametric equations for the line through the point (2, -8, 5) that is perpendicular to the plane 5x + 4y - 5z = 11. (Enter your answers as a comma-separated list. Let x, y, and z be functions of t.)

Solution

The line through (x0,y0,z0) that is parallel to the vector (a,b,c) has parametric equations

x = x0 + at, y = y0 + bt, z = z0 + ct.

You have the point (x0,y0,z0) = (2,-8,5).

The normal to the given plane is perpendicular to it hence parallel to the desired line. You can read this off of the coefficients of the plane. The vector is (5, 4, -5)---note these are the coefficients of x, y, and z, respectively. So the desired line is

x=2+5t

y=-8+4t

z=5-5t

Another approach:

Line will have equation:
(x, y, z) = (2, -8, 5) + t (a, b, c)

where vector (a, b, c) is normal to the plane.
Now plane Ax + By + Cz = D has normal (A, B, C)

So plane 5x +4y-5z = 11 has normal (5, 4,-5)

So line has equation (x, y, z) = (2, -8, 5) + t (5, 4,-5)

x=2+t

y=-8+4t

z=5-5t

are parametric equations of line.