"Find the equation of the plane that passes thru the pt (-4,-2,3) and contains the line (-4+4t,4-t,-2-t)."

Question

How would I go about doing this? I know how to find a plane with three points. I got two more points from the line, but when I tried them, I got the wrong answer.... Anybody know how to do this?

Answer 1

Find the equation of the plane that passes thru the point
P(-4,-2,3) and contains the line:
L = <-4 + 4t, 4 - t, -2 - t>

The directional vector u, of the line is:
u = <4, -1, -1>

A quick check shows that P is not on the line. This is good.

Let t = 0 and get a point on the line.
x = -4
y = 4
z = -2

The point is Q(-4, 4, -2). Now create a second directional vector for the plane v = PQ.

v = PQ = = <-4+4, 4+2, -2-3> = <0, 6, -5>

The normal vector n, to the plane is perpendicular to both directional vectors. Take the cross product.

n = u X v = <4, -1, -1> X <0, 6, -5> = <11, 20, 24>

With the normal vector and a point on the plane we can write the equation of the plane. Let's choose P(-4, -2, 3).

11(x + 4) + 20(y + 2) + 24(z - 3) = 0
11x + 44 + 20y + 40 + 24z - 72 = 0
11x + 20y + 24z + 12 = 0

Answer 2

The slope m of the line: m = distinction in Y-coordinates/ distinction of X-coordinates =(-2+6)/(one million-4) = - 4/3. The equation of the line is y-b =m(x-a). here, y+2 = (- 4/3)(x-one million) Or 3y +6 = - 4x +4. which finally ends up in : 4x+3y +2 =0. answer. confirm: ok

Answer 3

i always hated math..