How do I find the equation of a plane going through a given point and containing a line (given in paramentric)

Question

Here's my question:
Find the equation of the plane passing through the point P(4,0,-4) and containing the line
x=2-2t
y=1+7t
z=3+3t.

I cant seem to figure it out. thanks in advance for any help.

Answer 1

the direction vector of the line <-2 , 7 , 3> = A
a point on the line (2,1,3).
clearly, P is not on the line.

get the normal vector of the plane.... it is perpendicular to any vector lying on the plane.

one such vector is the direction vector of the line, A
another is obtained from the two points. call this B.

the normal vector is AxB.
You can now get the normal vector-point form of the plane.


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Answer 2

Find the equation of the plane passing through the point
P(4,0,-4) and containing the line

x = 2 - 2t
y = 1 + 7t
z = 3 + 3t

You need two directional vectors and one point to find the equation of the plane. We have the point P. The line contains one of the directional vectors u.

u = <-2, 7, 3>

We can get a second point Q, from the equation of the line by letting t = 0.

Q(2, 1, 3)

Now we can find a second directional vector v for the plane.

v = PQ = = <2-4, 1-0, 3+4> = <-2, 1, 7>

The normal vector n, of the plane is orthogonal to both u and v. Take the cross product.

n = u X v = <-2, 7, 3> X <-2, 1, 7> = <46, 8, 12>

Any non-zero multiple is also a normal vector. Divide by 2.

n = <23, 4, 6>

With the normal vector n and a point on the plane we can write the equation of the plane. Let's choose P(4, 0, -4).

23(x - 4) + 4(y - 0) + 6(z + 4) = 0
23x - 92 + 4y + 6z + 24 = 0
23x + 4y + 6z - 68 = 0

Answer 3

From the parametrics get two points, say t=0,1
x1, y1, z1 = 2,1,3
x2,y2,z2 = 0,8,6
Now set up the determinant

x y x 1
4 0 -4 1
2 1 3 1
0 8 6 1

Then plane Ax+By+Cz+D=0,
where A=0-41, B=-|4-41|,C=|401,D=|40-4
xxxxxxx 131| xxxx 231 xxxx 211xxx 213
xxxxxxx 861| xxxx 061xxxx 081xxx 046

Answer 4

placed the equation for the given line in slope-intercept (y=mx+b) style: 8x + y = 3 y = -8x + 3 Your "m" (slope) is -8. For a perpendicular line, take the destructive reciprocal slope and plug it right into a clean "y = mx + b" equation: y = (a million/8)x + b Now locate "b" by potential of utilising the given element: (x,y)=(3,6) 6 = (a million/8)3 + b 6 = 3/8 + b 5 5/8 = b = 40 5/8 Write the terrific form of the equation with m and b substituted in: y = (a million/8)x + 40 5/8