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.(adsbygoogle = window.adsbygoogle || []).push({});

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?????????

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# Parametric equation of a line and a plane

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