Parametric equation of a line and a plane

[linuspauling]

I CAN'T SEEM TO GET THE ANSWER THAT IS CONSISTENT WITH MY UNDERSTANDING OF THE USE OF DOT AND CROSS PRODUCTS AND THE USE OF THE PARAMETRIC EQUATION OF THE LINE.

LOOK AT THIS PLEASE:

the parametric equation of the line is:
x = 2 + 3t
y = -4t
z = 5 + t

the plane is 4x + 5y - 2z = 18

The simple part of course is to figure out the direction vector.

V = <3, -4, 1>

and if we use this basic vector fact, r = ro + vt where vt = a (vector)

r0 = <2, 0, 5>

this was the easy part.

I found that the plane and the line intersect when the parameter = -2, which gives us a point, p.

p(-4, 8, 3)

And, we all know that the NORMAL VECTOR of the given plane is
N = <4, 5, -2>

MY QUESTION FOR YOU...I NEED HELP!
SHOULDN'T THE CROSS PRODUCT BETWEEN THE VECTORS R0 AND THE DIRECTION VECTOR, V, GIVE US THE NORMAL VECTOR? OR AT LEAST A NORMAL VECTOR WITH COMPONENTS THAT ARE A MULTIPLE OF <4, 5, -2>?

WHAT ABOUT THE VECTOR THAT FORMS FROM THE POINT OF INTERSECTION OF PLANE AND LINE? VECTOR P = <-4, 8, 3>
IF I CROSS VECTOR P AND VECTOR R0, I OUGHT TO GET THE NORMAL VECTOR...BUT I DON'T!

WHAT PART OF MY UNDERSTANDING IS INCORRECT?

 

[fopc]

You already have an understanding of direction vector for line.
What you want now is a direction set for your plane.

Here's one way to compute them.
r1=(2,0,-5) is on the plane, and so is r2=(0,2,-4), and so is your p=(-4,8,3).
Claim: {r1-p, r2-p} constitutes a direction set for your plane (i.e., a linearly independent
doubleton set of direction vectors).

If you take the cross product of r1-p and r2-p, you should (will)
get a vector that's a scalar multiple of N.

How to "see" it? Visualize where the direction vectors reside.
If you cross them, what do you get?

 

[tacman]

It seems like you have a good understanding of the use of dot and cross products and the parametric equation of a line. However, your approach to finding the normal vector of the plane and the vector that forms from the point of intersection may not be correct.

Firstly, the cross product between the vectors r0 and v will not necessarily give you the normal vector of the plane. The normal vector of a plane is perpendicular to the plane itself, so it cannot be found by simply taking the cross product of two vectors within the plane. In fact, the normal vector of the plane is the coefficients of the x, y, and z terms in the plane's equation. In this case, the normal vector is <4, 5, -2>.

Secondly, the vector p that you have found is not necessarily the same as the vector that forms from the point of intersection of the plane and the line. The point p that you have found is the point where the line intersects the xy-plane, not necessarily the plane of interest. To find the point of intersection between the line and the plane, you would need to substitute the parametric equations of the line into the equation of the plane and solve for t. This will give you the value of t at which the line intersects the plane, and you can then substitute this value into the parametric equations to find the corresponding point of intersection.

In summary, to find the normal vector of a plane and the point of intersection between a line and a plane, you would need to use the equation of the plane and the parametric equations of the line, rather than taking the cross product of two vectors. I hope this helps clarify any misunderstandings you may have had.