Vectors and Planes

  • #1

Homework Statement


Question 1.
Given position vectors OA : i+2j-k and OB: 2i+j+3k in a 3D Cartesian space with origin O of the points A and B.

a) Find the scalar equation of the plane which contains A and which is perpendicular to vector AB.

b) Find the shortest distance from the point (1,-1,1) to the plane obtained in a (a)


Homework Equations





The Attempt at a Solution



Here is what i did
(a) Equation of line AB
r= (2,1,3)+ t(1,-1,4)

AP.n = 0
OP.n = OA.n
(x,y,z).(1,-1,4) = (2,1,3).(1,-1,4)

x-y+4z = 13 [equation of plane]?

(b) vector form of plane => r.(1,-1,4) = 13
line equation => r= (1,-1,1) + t(1,-1,4)

Using r=r for both equations i get t= 2/9

Substitute t=2/9 into line equation giving me (11/9, -2/9, 17/9)

I'm stuck here then for part (b)
I'm pretty weak in this chapter, couldn't seem to grab the concept here.

Homework Statement


Question 2.
Obtain the scalar equation of the plane which passes through point P(1,2,3) and contain a straight line
x(t)= 3t
y(t)= 1+t
z(t)= 2-t
t is the parameter of the line


Homework Equations





The Attempt at a Solution



Equation of the line= (0,1,2) + t(3,1,-1)

Let Q and R be the points on the line
t=0 Q=(0,1,2)
t=1 R= (3,2,1)

Therefore PQ= (-1,-1,-1) PR= (2,0,2)

PQxPR= (-2,0,-2)

Thus the equation of the plane
a(x-x1) + b(y-y1) + c(z-z1)= 0
-2(x-1) + 0 -2(z-3) = 0
-2x - 2z + 8 = 0
is this correct?
 
Last edited:

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770

Homework Statement


Question 1.
Given position vectors OA : i+2j-k and OB: 2i+j+3k in a 3D Cartesian space with origin O of the points A and B.

a) Find the scalar equation of the plane which contains A and which is perpendicular to vector AB.

b) Find the shortest distance from the point (1,-1,1) to the plane obtained in a (a)


Homework Equations





The Attempt at a Solution



Here is what i did
(a) Equation of line AB
r= (2,1,3)+ t(1,-1,4)

AP.n = 0
OP.n = OA.n
(x,y,z).(1,-1,4) = (2,1,3).(1,-1,4)

x-y+4z = 13 [equation of plane]?

You will find that (1,2,-1) doesn't work in that equation so the point is not on the plane.
For R = <x,y,z>, the form on the equation of a plane is

[tex](\vec R - \vec A)\cdot \vec N = 0[/tex]

The correct equation for the plane might get you started.
 
  • #3
so what does it mean when it says the plane contains A?

so is it like this?
[(x,y,z)-(1,2,-1)].(1,-4,4)=0

1(x-1)-4(y-2)+4(z+1)=0

x-4y+4z=11
 
Last edited:
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
so what does it mean when it says the plane contains A?

so is it like this?
[(x,y,z)-(1,2,-1)].(1,-4,4)=0

1(x-1)-4(y-2)+4(z+1)=0

x-4y+4z=11

Yes, except check the sign on the 11.
 

Related Threads on Vectors and Planes

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
19
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
2
Views
716
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
875
Replies
2
Views
8K
  • Last Post
Replies
3
Views
3K
Top