# Vectors and Planes

## 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)

## 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

## 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:

LCKurtz
Science Advisor
Homework Helper
Gold Member

## 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)

## 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.

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:
LCKurtz
Science Advisor
Homework Helper
Gold Member
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.