Vector Problem help pls?

Question

relative to origin O

OA = (-1, 3, 5) OB = (3, -1, -4)

ON = ( 5, 1, -3)

The line L passes through A and is parallel to OB. the point N is the foot perpendicular from B to L


L = (-1 , 3, 5) + t ( 3, -1, -4)

length BN is 3.

how to find the equation of the plane containing A, B & N?


Thanks

Answer 1

You need two things to find an equation of a plane.
1. A point in the plane
2. A normal vector to the plane.

You already have a point in the plane (3 of them in fact) so you need to find the normal vector. To find the normal vector you take the cross product of two nonparallel vectors that are in the plane. For instance, find AB and AN. Then take their cross-product to get the normal vector.

Answer 2

There seems to be some irrelevant information. I assume the question is simply to find the equation of the plane thru the points A, B, and N.

First from two directional vectors for the plane from the three points.

A(-1, 3, 5); B(3, -1, -4); and N(5, 1, -3)

AB = = <3+1, -1-3, -4-5> = <4, -4, -9>
AN = = <5+1, 1-3, -3-5> = <6, -2, -8>

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

n = AB X AN = <4, -4, -9> X <6, -2, -8> = <14, -22, 16>

Any non-zero multiple of n is also a normal vector to the plane. Divide by 2.

n = <7, -11, 8>

With the normal vector to the plane and a point on the plane we can write the equation of the plane. Let's choose
A(-1, 3, 5).

7(x + 1) - 11(y - 3) + 8(z - 5) = 0
7x + 7 - 11y + 33 + 8z - 40 = 0
7x - 11y + 8z = 0

Answer 3

This comment plays with cross products and vectors about a plane, the answer is within somwhere, not necc. in it's final form, in which unit vector n (normal to plane) is used

A(a1,a2,a3) and B(b1,b2,b3) an N(n1,n2,n3)

with [a,b,c] being a vector normal to the plane
the plane passes through a point (x0,y0,z0)
equation of the plane is
a(x-x0), b(y-y0), c(z-z0)=0

to find equation using A B and N:
find a vector, on the plane, by taking the difference between two points on plane
[x0,y0,z0] = B-A = [Bx-Ax, By-Ay, Bz-Az]
[3- -1, -1-3, -4-5] = [4, -4, -9]

(x,y,z) = N-B = [5-3,1- -1, -3-4] = [2,2,-7]

===========================================
cross product of these two vectors that lie on plane =
a line perpendicular to the plane
cross product = [a2b3-a3b2, a3b1-a1b3, a1b2-a2b1]
cross [B-A] with [N-B]= [46,10,16] =[a,b,c]
=========================================
dot product of cross product and vector on plane should =0
[a,b,c]dot[B-A] = a(Bx-Ax), b(By-Ay), c(Bz-Az)

= a(4), b(-4), c(-9)=0 Walla equation for plane

carried out with normal vector 46, 10, 16 (is this vector a unit vector? I don't think so, but it is orthogonal to the plane!)
46(4),10(-4), 16(-9) = 184-40-144= 0

now you have shown yourself it makes sense somewhat
so going back
if you need an expression with AB and N, show the dot product we just solved but with B-A and N-B as the points


a((B-A)x-(N-B)x)+((B-A)y-(N-B)y)+
((B-A)z-(N-B)z)=0

this plane is not with respect to the origin

if just needing equation for plane with A B and C
then ===================================
a(Bx-Ax)+ b(By-Ay)+ c(Bz-Az)=0
a(4)+ b(-4)+ c(-9)=0
==================================

Answer 4

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