# Cartesian equation of a plane

• April 13th 2010, 10:20 PM
Dranalion
Cartesian equation of a plane
I am stuck on this problem!

(a) Find the cartesian equation of the plane P through the points
$P(1, 2, 1)$, $Q(2, 1, -1)$, $R(0, 3, 1)$.

(b) Find the equation of the line which is perpendicular to P and passes through the point $S(2, -2, 3)$.

(c) Find the coordinates of the point T on P which is closest to S and hence find the distance from S to P.

Any help is much appreciated,

Dranalion
• April 13th 2010, 10:54 PM
Dranalion
For (a), I found the vectors PQ and PR.

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

Cross product of PQ and PR is: 2i +2j +2k = (2, 2, 2)

Therefore the equation of the plane is:

P = 2x + 2y + 2z? (Is this correct?)

Or should I substitute the point P(1, 2, 1) into the equation 'P' to get:
2(1) + 2(2) + 2(1) = 8

Therefore, 2x + 2y + 2z = 8

Which is correct? I am completely stuck for parts (B) and (C)!
• April 13th 2010, 11:12 PM
Rapha
Hi.

Quote:

Originally Posted by Dranalion
For (a), I found the vectors PQ and PR.

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

Erm, no?

I get PQ = (1,-1,-2). Not '+2'

Quote:

Originally Posted by Dranalion
Cross product of PQ and PR is: 2i +2j +2k = (2, 2, 2)

It's a good idea to calulate the cross product,

I calculated

(1,-1,-2)x(-1, 1, 0) = (2,2,0)

Quote:

Originally Posted by Dranalion
Therefore the equation of the plane is:

P = 2x + 2y + 2z? (Is this correct?)

No, because ...

Quote:

Originally Posted by Dranalion
Or should I substitute the point P(1, 2, 1) into the equation 'P' to get:
2(1) + 2(2) + 2(1) = 8

Therefore, 2x + 2y + 2z = 8

Yea, exactly. Now you have the information, that the point P is in the plane.

If you substitute the points Q and R as well, the left side should be equal to 8.

As you can see it is wrong if you substitute R in this equation

Quote:

Originally Posted by Dranalion
Which is correct? I am completely stuck for parts (B) and (C)!

(B) You already know the euclidean vector of the line - it is (2,2,0) (you calculated (2,2,2) instead...) . This is the vector you should use, because you know it is orthogonal/perpendicular to the plane p

So the line you need is

L: OP + t*(2,2,0)

If you substitute t = 0 the line L goes through P.

(C) Of course the distance got to be 0, because the line is not parallel to the plane.
So you can calculate the intersection point of L and P or you can use the 'hesse normal form'.

Whatever you like.

Rapha
• April 13th 2010, 11:36 PM
Dranalion
Quote:

Originally Posted by Rapha
(B) You already know the euclidean vector of the line - it is (2,2,0) (you calculated (2,2,2) instead...) . This is the vector you should use, because you know it is orthogonal/perpendicular to the plane p

So the line you need is

L: OP + t*(2,2,0)

If you substitute t = 0 the line L goes through P.

(C) Of course the distance got to be 0, because the line is not parallel to the plane.
So you can calculate the intersection point of L and P or you can use the 'hesse normal form'.

Whatever you like.

Rapha

Sorry, I don't understand!
What do you mean?
• April 14th 2010, 12:07 AM
Rapha
Quote:

(b) Find the equation of the line which is perpendicular to P and passes through the point http://www.mathhelpforum.com/math-he...ebfd95d1-1.gif.
Quote:

Originally Posted by Dranalion
Sorry, I don't understand!
What do you mean?

In (b) you want to find a line which is orthogonal to P and goes through the point S(2,-2,3)

A line has the equation

L: 0X + t XY, where X and Y are coordinates and t is a real number.

So

$L: 0S + t XY = \begin{pmatrix} 2-0 \\ -2-0 \\3-0 \end{pmatrix}+ t XY$

If you substitute t = 0 in this equation, then

$\begin{pmatrix} 2 \\ -2 \\3 \end{pmatrix} + 0* XY = \begin{pmatrix} 2 \\ -2 \\3 \end{pmatrix} =S$

Hence every line with this position vector 0S apparently goes through the point S.

And what you now want to do is finding a vector XY such that L is orthogonal to the plane P.
To find an orthogonal vector you need to calculate the cross product.

so if XY = (2,2,0) we get

L: (2,-2,3) + t*(2,2,0)

so this line is perpendicularto the plane P, because the vector (2,2,0) is orthogonal to the plane P. (This is what the normal vector [of the plane P] is about)

Are there still any questions on (b)?