If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Normal vector from plane equation

Figuring out a normal vector to a plane from its equation. Created by Sal Khan.

## Want to join the conversation?

• This video explains how to get a normal vector, but a plane has infinite number of normal vectors right? I mean, each point on the plane has a normal vector. Is the normal vector calculated in this video special somehow? or why is the other normal vectors not explicitly written in the plane equation?
• You're correct in that each point will have a normal vector, however since we're dealing with a flat plane each of those normal vectors will be the same (they will all point in the same relative direction). Therefore all the points on the plane share a "unique" normal vector.
If instead we were dealing with a sphere, or some other non-flat surface, then the normal vector would be different at each point on the surface and we would have a similar situation to what you were describing.
• I still don't understand why A, B, and C are the normal vector coordinates..
• Here's how I like to think of it. I've got a line in R^2
Ax+By=C.
If I want to find a normal vector, I can find the slope of the line and then do the opposite reciprocal to find a normal vector.
By=-Ax+C
y=-A/B*x+C/B. The slope is -A/B. A normal vector will have slope B/A. An easy way to construct this is to make the y comp = B and the x comp = A. Thus, the vector normal the line Ax+By=C is [A, B]. The jump to 3D is now more intuitive.
• Thanks for the great tutorials Sal.
I'm learning, however I'm wondering if you have the y and z axes correct?
In this video Normal vector from plane equation - , you label the x-axis as right (correct), y-axis as front (should be up) and z-axis as up (should be front).
I tried turning my head so x-axis points right (correct) and y-axis points up (correct), however then the z-axis points to the back, meaning the positive z you wrote, should be negative z.
Anyway, I'm learning so I might be wrong, however I would like to be clear on this as others I have spoken to seem unsure.
• Yes, the coordinate system Sal made is left-handed, when usually a right-handed system is used.
• How come this section doesn't have any quizzes?
• Maybe since the topics are quite diverse and quizzes already exist in the other sections
the team won't mention them here - maybe the IIT JEE questions are a bit too difficult.
• Why is the normal vector equation n.(ai + bj +ck) ?
• I, j and k are the unit vectors towards the three different axes. All vectors in our three dimensional space can be expressed using them.
The advantage of i, j and k come from the fact that they are all perpendicular to each other. This makes calculating them a lot of easier.
Hope I helped.
• I don't see how P-P1 can "lie along the plane". You've subtracted two position vectors - surely this would just give a third position vector, ie define a point? I can see that the line from the origin to the position defined by this third position vector would be in a second plane parallel to the first plane, but the second plane must include the origin. Or am I misunderstanding something?
• Sal says P-P1 lies "on" the plane because it does when started from any point on the plane (as Sal did in his example). Drawing it between the endpoints of P and P1 helps illustrate why P-P1 is part of defining the plane, which is why he did so in his earlier video and again now.

However, you are correct that vectors can be moved anywhere, so implying one is always on a specific plane is misleading. A more correct statement would be that P-P1 lies on a plane parallel or equal to the one Sal is using. But since these planes all have the same normal vectors, that distinction isn't necessary for what Sal's trying to explain now.
• at he has expanded the dot product of (P-P1) DOT N

P-P1 is (X-Xp)i + (Y-Yp)j + (Z-Zp)k
N is Ai + Bj + Ck

But he ignores the i, j, and k during his expansion:

AiX- AiXp + BjY - BjYp + CkZ - CkZp = 0
He says
AX - AXp + BY - BYp + Cz - CZp = 0

Why does he lose out the unit vectors in the dot product expansion?
• Because the dot product takes as input two vectors, but returns a scalar number as result, remember that one way to look at the dot product is: `A ∙ B = |A| |B| cos(θ)` where `θ` is the angle between vectors `A` and `B`.
• Hi Sal. Thanks for your great videos.
Question: (0.30) why do we describe the vector n = ai + bj + ck and not n = (ai, bj, ck)?
If i,j,k are representing the three axis of a plane, why are we adding them together instead of defining them as a tuple?
• a1i + a2j + a3k
= a1(1, 0, 0) + a2(0, 1, 0) + a3(0, 0, 1)
= (a1, 0, 0) + (0, a2, 0) + (0, 0, a3)
= (a1, a2, a3).
• Is D the distance of the plane from the origin ?
• Kinda. Let's work in 2D. let's say you have x + y = 2. So here A and B = 1 This is just another way of saying y = 2 - x. In this case D = 2, and it moves the line up 2 units. So yes, where x = 0 the line is 2 above the origin, but elsewhere along the line it won't be 2 away. the 2 just tells you how far upward you have to move the line y = -x

More generally let's say Ax + By = D, now we will solve for y. y = D/B - A/Bx. So now we have the line -A/Bx being moved along the y axis D/B units, so still, the D entry determines how far along the y axis the line is moved. Or at least it plays a big part. if D = 0 then the line would cross through the origin.

Similarly in 3d D tells you how far along the z axis you have to move the plane, or at least it plays a major part. So starting with Ax + By + Cz = D and solving for z we get (D - Ax - By)/C = z And, if you plug in (0,0) for x and y, you will be exactly D/C away from the origin along the z axis. And if D was 0 you would have the plane (-Ax - By)/C = 0 which goes through the origin.

It's worth noting you can solve for any of the three variables and still get the answer. Let's use 3x + 5y = 7 and hopefully you'll be able to see how it follows in 3d. So, let's solve that for x and y, in both cases you get:

(7 - 5y)/3 = x
(7 - 3x)/5 = y

Try graphing both and you should see you get the same graph. use a site like desmos or geobra if you need. But in the y= one the 7/3 term tells you how far you go up the y axis, while in the x= equation the 7/5 tells how far you move the line along the x axis.

I really hope this helpped, the main takeaway is that the D term plays a major part in telling how far a line/ plane is moved away from the origin.
• So, basically, a normal vector to the plane Ax+By+Cz=D is [A, B, C]?

If this is true, one could find the equation of a plane by knowing the normal vector and 1 point in a very straight forward way. Ax+By+Cz is known from the normal vector and D can be found by putting the coordinates of the point in. Very useful!

For example, let's say [3, 1, -1] is the normal vector and (2, 1, 4) is a point on the plane. I instantly know that Ax+By+Cz=3x+y-z. D is found using the point. D=3x+y-z=3*2+1-4=3. This result in the plane 3x+y-z=3.

Is this a proper way of solving an exercise?