Equations of a Plane

Transcript

0:00Introduction

Hello everyone! Put a stool with three legs on any floor, and it never wobbles. But a table with four legs often rocks on the same floor. Three legs doing better than four? Strange, right? The reason is pure geometry, and you will see it today. Planes come up often in JEE, so let's begin.

0:20One point and one normal fix a plane

Suppose I fix one point, and one direction perpendicular to the plane, called the normal. Now tell me, is the plane fixed, or can it still move? Yes, exactly, it is fixed! Through the point x one, y one, z one, with normal a, b, c, the plane is: a times, x minus x one, plus b times, y minus y one, plus c times, z minus z one, equals zero. The small ones are subscripts. It works because every vector in the plane is perpendicular to the normal. Okay? Open the brackets, and you get the general form: a times x, plus b times y, plus c times z, plus d, equals zero. The coefficients of x, y and z give the normal straight away. Clear?

1:12Normal form: a unit normal and a distance

Now a special form, the normal form: vector r dot n hat equals d. The arrow means a vector, the dot is the dot product, and the cap on n, called a hat, means a unit vector. So n hat is the unit normal, and d, greater than or equal to zero, is the distance of the plane from the origin, O, okay? In Cartesian form, that is in coordinates, it reads l times x, plus m times y, plus n times z, equals d. Here l, m, n are the direction cosines of n hat, written as d c's on screen, so l squared plus m squared plus n squared equals one. Got it? Now a common trap. In the general form, is the constant d already the distance from the origin? Not always, so be careful! It is the distance only for a unit normal. So first divide the equation by the square root of a squared plus b squared plus c squared, and keep the constant on the right side not negative, okay?

2:15Intercept form: where the plane meets the axes

Next, the intercept form: x over a, plus y over b, plus z over c, equals one. Careful, here a, b and c are not the normal at all, right? So tell me, where does this plane meet the x axis? On that axis, y and z are both zero. Yes, x over a equals one, so x equals a. So the plane meets the axes at a, zero, zero, then zero, b, zero, and zero, zero, c. These numbers a, b, c are the intercepts. Clear?

2:56Through three points: the cross product gives the normal

Now, back to our stool. Its three feet are three points: A, B and C, okay? The vectors A B and A C both lie in the plane of the floor. So which vector is perpendicular to both? Yes, exactly, the cross product! The normal is vector A B cross vector A C, where the arrow over A B means the vector from A to B, and the cross sign means the cross product. Right? The same thing as a determinant, where the vertical bars mean a determinant, and it equals zero. Point A is x one, y one, z one, and B and C carry the subscripts two and three. The first row is x minus x one, and so on, then come B minus A and C minus A, got it? And one condition: the points must not lie on one straight line, otherwise the cross product is the zero vector, the zero with an arrow. That is the stool's secret! Three feet not in a row always sit in one plane, but a fourth foot need not. Clear?

4:00Planes from lines: cross two vectors in the plane

Now lines, with the same rule: find two vectors in the plane, and cross them. Take a line through the point with position vector a, along vector b, and a point P off it. Both b, and P minus vector a, lie in the plane. So the normal is b cross, P minus vector a, okay? Two coplanar lines, which means lying in one plane, and not parallel: the normal is b one cross b two, their two directions. Right? But for two parallel lines, use b cross, a two minus a one, the vector joining a point of one line to a point of the other. Example two will show you why. And through a point, parallel to u and v, the normal is simply u cross v. Four cases, one idea. With me?

4:48Example 1: plane through (1,1,0), (1,2,1) and (-2,2,-1)

Okay, let's solve one, and just like our stool, it starts with three points. Find the plane through the three points: one, one, zero; one, two, one; and minus two, two, minus one. Pause the video and try it first. Done? Okay, let's check. Call the points A, B and C in that order. Vector A B is zero, one, one, and vector A C is minus three, one, minus one. Okay? Now, what is vector A B cross vector A C? Watch the middle part, it carries a minus sign in front. Let's work it out: the first part, one times minus one, minus one times one, gives minus two. The middle part, with its minus sign, gives minus three, and the last part gives three. So the normal is minus two, minus three, three. Now use A, which is one, one, zero: minus two times, x minus one, minus three times, y minus one, plus three times, z minus zero, equals zero. Open the brackets and multiply by minus one; the double arrow means implies. So, two x plus three y minus three z equals five. Now check with C, the point we did not use: two times minus two, plus three times two, minus three times minus one. That is minus four plus six plus three, which gives five. It works perfectly, so the plane is right.

6:20Example 2: plane containing two parallel lines

Now a harder one. Find the plane containing the two parallel lines: x minus one, over one, equals y minus two, over two, equals z minus three, over three; and x minus two, over one, equals y plus one, over two, equals z over three. Pause the video and try it first. Done? Okay, let's check. Both lines have direction ratios one, two, three. So, shall we take b one cross b two as the normal, as before? No, no, careful, that is the trap! Equal directions give the zero vector as their cross product, and it has no direction, so it cannot be a normal. Okay? So we need a vector joining the two lines. Vector a one is one, two, three, and vector a two is two, minus one, zero. Their difference is one, minus three, minus three. Right? Now cross one, two, three with one, minus three, minus three. The first part, minus six plus nine, gives three. The middle part, with its minus sign, gives six. The last part, minus three minus two, gives minus five. So the normal is three, six, minus five. Through one, two, three: three times, x minus one, plus six times, y minus two, minus five times, z minus three, equals zero. The constants cancel, so the plane is three x plus six y minus five z equals zero. Now check the second line's point, two, minus one, zero. Three times two, plus six times minus one, minus five times zero, gives zero. So both lines lie in the plane, very good!

8:09Quick revision: point and normal, normal form, cross product

So, quick revision. One point and one normal fix a plane, with this equation. In the normal form, d is the distance only with a unit normal. And with no normal given, cross two vectors in the plane; for parallel lines, use a two minus a one. Now try the practice problems on this page. All the best!