Scalar Plane Equation for Vectors AB and Point P | 3D Cartesian Space

[speedycaster]

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?

 

[LCKurtz]

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

(\vec R - \vec A)\cdot \vec N = 0

The correct equation for the plane might get you started.

 

[speedycaster]

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

 

[LCKurtz]

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.