# Find plane, given a line & perpendicular plane

• Excelled
In summary, the problem is to find an equation of a plane that passes through the line x+y=2, y-z=3, and perpendicular to the plane 2x+3y+4z=5. To solve this, one must be able to find a vector in the plane which is perpendicular to the given plane and also lie in that plane. Once this is done, solving for the vector will give the equation of the plane.
Excelled
Please help me! I have been sitting with this problem for god knows how long, and I just can't figure it out. I've tried re-reading the problem text, re-reading the chapter, reading alternative explanations on the web, drawing the problem on paper -- heck, I've even tried shouting at it -- but no luck. Can someone please give me some pointers?

The problem?
Find the equation of the plane that satisfies the given conditions:
Passing through the line x+y=2, y-z=3, and perpendicular to the plane 2x+3y+4z=5.

Firstly, I'm not even sure if I'm reading it correctly. Does passing through mean that the plane contains the line, or that it passes through a point in the line? Is it one line (x+y=2, y-z=3) or two lines (x+y=2 and y-z=3)?I haven't seen it in that form before. Is it the wanted plane or the line that is perpendicular to the given plane? Secondly, how do I solve it?

I have a feeling that this should be easy, so this is really bad for my self-confidence. :-(

Last edited:
Excelled said:
Please help me! I have been sitting with this problem for god knows how long, and I just can't figure it out. I've tried re-reading the problem text, re-reading the chapter, reading alternative explanations on the web, drawing the problem on paper -- heck, I've even tried shouting at it -- but no luck. Can someone please give me some pointers?

The problem?
Find the equation of the plane that satisfies the given conditions:
Passing through the line x+y=2, y-z=3, and perpendicular to the plane 2x+3y+4z=5.

Firstly, I'm not even sure if I'm reading it correctly. Does passing through mean that the plane contains the line, or that it passes through a point in the line? Is it one line (x+y=2, y-z=3) or two lines (x+y=2 and y-z=3)?I haven't seen it in that form before. Is it the wanted plane or the line that is perpendicular to the given plane? Secondly, how do I solve it?

I have a feeling that this should be easy, so this is really bad for my self-confidence. :-(
Since just being told a line that passes through the plane and that it is perpendicular to another plane would not be enough to identify the plane, this must mean that the line lies in the plane- every point of the line is in the plane. And "x+ y= 2, y- z= 3" gives a single line, not 2. In 3 dimensions, a single equation is not sufficient to define a line. In effect, a single equation defines a plane. Saying "x+ y= 2, y- z= 3" means that the line is the intersection of those planes.
To find a plane, it is sufficient to determine a single point in the plane, (a, b, c), and a vector, Ai+ Bj+ Ck, perpendicular to the plane. Then the equation of the plane is A(x-a)+ B(y-b)+ C(z-c)= 0. In particular, the plane 2x+3y+4z=5 is perpendicular to the vector 2i+ 3y+ 4k. Since the plane you seek is perpendicular to the given plane, that perpendicular, 2i+ 3y+ 4k, must be in the plane. Knowing that the line x+ y= 2, y- z= 3 is in the plane tells us that a vector in its direction must also be in the plane. We can write x= 2- y, z= y- 3 so we can take y itself as parameter: parametric equations for this line are x= 2- t, y= t, z= t- 3. The coefficients for x, y, z, -1, 1, 1, respectively, tell us that the vector -i+ j+ k is in the direction of the line and so also in the plane. You now know two vectors in the plane and you should know that their cross product is perpendicular to the plane itself. Now you know a vector perpendicular to the plane. All you need is a single point in the plane which you can get by plugging any value of t into the parametric equations for the given line.

Thank you so much for that great explanation, HallsofIvy! It made it clear for me. I can finally continue. :-)

## 1. What is the equation for finding a plane given a line and a perpendicular plane?

The equation for finding a plane given a line and a perpendicular plane is known as the "point-normal form" equation, which is:

ax + by + cz = d

Where (a,b,c) is a vector perpendicular to the plane, and (x,y,z) is a point on the plane.

## 2. How do you determine the values of a, b, c, and d in the point-normal form equation?

First, you will need to find two vectors that lie in the plane. Then, you can use the cross product of these two vectors to determine the coefficients a, b, and c. Finally, you can plug in the coordinates of any point on the plane to find the value of d.

## 3. Can you find the equation of a plane if only given a line and a point on the plane?

Yes, you can use the "point-slope form" equation to find the equation of a plane given a line and a point on the plane. The equation is:

ax + by + cz = d

Where the coefficients a, b, and c are determined by the direction of the line, and d is the distance from the origin to the plane.

## 4. How do you graph a plane given its equation in point-normal form?

To graph a plane given its equation in point-normal form, you can first find three points that lie on the plane by plugging in different values for x, y, and z. Then, plot these points on a 3D coordinate plane and connect them to create the plane.

## 5. Are there other methods for finding a plane given a line and a perpendicular plane?

Yes, there are other methods for finding a plane given a line and a perpendicular plane. One method is to use the "cross product form" equation, which is:

l(x-x0) + m(y-y0) + n(z-z0) = 0

Where (l,m,n) is a vector perpendicular to the plane, and (x0,y0,z0) is a point on the plane. Another method is to use vector projections to find the equation of the plane.

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