Find the parametric equations for the line through the point P = (5, -5, 5) that is perpendicular to the plane 4x+5y−2z=1.

Question

 any help is appreicated

Answer 1

(A) The perpendicular from point P(5,-5,5) is parallel to the normal vector for the plane. The parametric equations take the form:

x=x₁+at, y=y₁+bt, z=z₁+ct 

where x₁, y₁, and z₁ are the coordinates of P: 5, -5, and 5 respectively.

The normal vector n=<4,5,-2> being the coefficients of x, y and z in the equation of the plane. The perpendicular through P is parallel to the normal so we can use the normal vector to create the parametric equations: <a,b,c>=<4,5,-2>, giving us the equations:

x=5+4t, y=-5+5t, z=5-2t

Any scalar multiple of these produces a line perpendicular to the given plane and passing through P.

(B) We can substitute for x, y, z in the equation of the plane:

4x+5y-2z=1 becomes 4(5+4t)+5(-5+5t)-2(5-2t)=1,

20+16t-25+25t-10+4t=1,

45t-15=1, 45t=16, t=16/45.

Therefore, x=5+64/45=289/45, y=-5+16/9=-29/9, z=5-32/45=193/45, giving us the point Q(289/45,-29/9,193/45). This is where the perpendicular meets the given plane.

If Q is the point in the y-z plane rather than on the given plane then x=0=5+4t, so t=-5/4. The corresponding y and z values are y=-5+5(-5/4)=-45/4, z=5-2(-5/4)=15/2, so Q would be (0,-45/4,15/2). P actually lies between the given plane and the y-z plane.