# Homework Help: Finding distance of "P" from a plane OQR using vector

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1. Oct 19, 2017

### LeitHunt

1. The problem statement, all variables and given/known data
Find th distance of point P to the plane OQR, O being the origin.
"PS" is perpendicular to plane "OQR"
P(3,-2,-1)
S(x,y,z)

O(0,0,0)
Q(1,3,4)
R(2,1,-2)

2. Relevant equations
A) The equation in terms of (x,y,z) obtain by triple product of OS,OQ & OR.
OS.(OQXOR)=0......... [all being co-planar therefore =0]
we get,
-10x+10y-5z=0

B)The Equation on which I'm stuck.I didn't get how PS vector is obtained from coordinates of P & S ;PS=(3-x)i+(-2-y)j+(-1-z)k

Then direction ratio is used to get
x+y=1
y+2z=-4

Now we have 3 equations & 3 Unknowns
3. The attempt at a solution
I understood how first equation is obtained that means by using theorem :- three vector being co-planar then
A.(BXC)=0..... (as we get volume from this, being co-planar means no volume)
So equation 1 is obtained that way

Now in Second equation
Vector cross product is done OQ X OR= -10i + 10j -5k
Then,
Where I'm stuck:-
Vector PS is given by (3-X)i + (-2-Y)j+(-1-Z)k,
How??? Vector PS is obtained by coordinate of P & S????
Next two equations are simple it just uses the direction ratio.
Solving 3 equation I'll get x,y& z.
Then find vector PS from above and find magnitude of PS that means it's length.
For reference please check below image :-
https://m.imgur.com/a/dEXAT

2. Oct 20, 2017

### ehild

$\vec {PS} = \vec {PO} - \vec {SO}$
You make it too complicated. Finding the normal unit vector $\vec n$
of the plane OQR, the distance of point P from the plane is the projection of $\vec{PO}$ onto $\vec n$.

3. Oct 20, 2017

### LeitHunt

I understood now about PS :).

1.But how to find a unit vector of plane in this case?

2.Also to find projection of OP I'll need angle POS but "S" co-ordinates are not known. I'm struggling to understand/imagine which angle will be needed for finding component of OP :(

3.[Edit:- I think I get this one,here P is defined from origin O not from S so vector PS is not (3i-2j-k) it's vector OP.]----->I know this is wrong but I have some confusion :-
Why can't I just use vector SP obtained from p(3,-2,-1) find it's magnitude which will be equal to distance of P from plane.Why it's wrong?

Thank you :) for helping

Last edited by a moderator: Oct 20, 2017
4. Oct 20, 2017

### ehild

You need the unit normal vector of the plane OQR. You get a normal vector as the cross product of vectors OQ and OR, and dividing it by the norm, it is the normal unit vector pf the plane.

The angle POS is the same as the angle the vector $\vec{OP}$ encloses with the normal vector $\vec n$, see picture.

It is not wrong, but complicated.
Point P defines the position vector $\vec {OP} = 3i -2j -k$
The scalar product $\vec {OP} \cdot \vec n= |OP| |n| \cos(\theta) = |OP| \cos(\theta)$ , is the projection of $\vec {OP}$ on the direction of the normal vector, and it is the distance of point P from the plane OQR.

5. Oct 22, 2017

### LeitHunt

Ok now everything is clear. Thanks