Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have been working through all the problems in my textbook and I have finally got to a difficult one. The problem ask

Find the equation of the plane.The plane that passes through the points $(-1,2,1)$ and contains the line of intersection of the planes $x+y-z =2$ and $2x-y+3z=1$

So far I found the direction vector of the line of intersection to be $<1,-2,1>$ and I have identified a point on this line when $x=0$ to be $(0,5,-1)$.

I do not know how to find the desired plane from here.

Any assistance would be appreciated.

share|improve this question
    
Do you know how to find the equation of a plane given three points that lie in it? Or two vectors and a point? –  orlandpm Jan 18 '13 at 4:10

3 Answers 3

up vote 1 down vote accepted

Here is a related problem. To find the equation of the plane, you need three points which lie in the plane. Already, you have one, you can get the other two from the equation of the line, since the line lies in the plane. Once you have the three points, construct two vectors say $v_1$ and $v_2$ and find the cross product $n=v_1 \times v_2$. Then, the equation of the plane is given by

$$ n.((x,y,z)-(-1,2,1))=0 .$$

share|improve this answer

The following determinant is another form of what Scott noted: $$\begin{vmatrix} (x-x_0)& (y-y_0)& (z-z_0)\\1& 1& -1\\2& -1& 3 \end{vmatrix}$$

share|improve this answer
    
I have determined that the cross product is <2,-5.-3>. How do I determine the <a,b,c> in the equation of the plane? –  jascal Jan 18 '13 at 4:34
    
@jascal: if you did it correctly, yes it was. ;-) –  Babak S. Jan 18 '13 at 4:35
    
So my answer would be 2(x+1)-5(y-2)-3(z-1)=0, however, the answer in the back of the book is -2(x+1)+4(y-2)-8(z-1)=0 –  jascal Jan 18 '13 at 4:38
    
@jascal: Did you get yor answer? can you find the equaton? –  Babak S. Jan 18 '13 at 7:22
    
I like this explanation! +1 –  amWhy Feb 14 '13 at 0:12

To find a plane in space, you need a normal direction vector (A,B,C) and a point. The equation is $A(x-x_0) + B(y-y_0) + C(z-z_0)=0$. In your case, that is $A(x+1)+B(y-2)+C(z-1)=0.$ Now take the cross product of the normals to your other planes $(1,1,-1)\times(2,-1,3)$ to get a vector that is perpendicular to the line of intersection.

share|improve this answer
    
I have determined that the cross product is <2,-5,-3>. Is this the <a,b,c> in the equation of the plane? –  jascal Jan 18 '13 at 4:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.