Vector Methods in Geometry

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Draw any parallelogram you like, and then draw its two diagonals. Where do they cross? Exactly at the middle of both, every single time, right? But how do you prove it? In school geometry, you need congruent triangles and a page of angles. Today we will prove it in a few short lines, with vectors, and with no angles at all. Questions like this come up often in JEE vector algebra. So, let's begin.

0:33The standard method in four moves

So, here is the standard method, in four moves. Move number one: choose an origin, and very often, we choose a vertex of the figure. Why a vertex? Because the origin has position vector zero, so that point simply drops out of the working. Nice, right? Move number two: name two or three independent vectors, usually the sides that start from that vertex. Independent means two vectors that are not collinear, or three vectors that are not coplanar. Okay? Move number three: write every point in terms of them, the vertices, the midpoints, the points that divide a side, everything. Clear? Now suppose the same point comes out in two different ways, say, from two different lines. What do we do then? Yes, exactly, we set the two expressions equal and compare coefficients, the numbers in front of each vector. That gives simple equations, and the equations give the answer. With me?

1:33When can we compare coefficients?

Now, why are we allowed to compare coefficients? Take two vectors, a vector and b vector, with the small arrows on top, and let them be non collinear. Then every vector in their plane can be written as x times a vector, plus y times b vector, and in exactly one way. Okay? So if x times a vector plus y times b vector equals p times a vector plus q times b vector, then x equals p, and y equals q. The double arrow means implies. Right? And in space, with three non coplanar vectors a, b and c, the same is true for x times a vector, plus y times b vector, plus z times c vector. Got it? Now let me ask you this: what goes wrong if the two vectors are parallel, say, b vector is two times a vector? Then two a vector plus zero b vector, and zero a vector plus one b vector, are the same vector, and yet the coefficients are different! So, careful: compare coefficients only when the vectors are independent. That is exactly why move two says independent. Clear?

2:39Tools for writing every point

Now, move three needs a few tools from earlier lessons. In the figure, O is the origin, and the arrows a and b are the position vectors of the points A and B, okay? The long arrow over A B means the vector from A to B, and it equals b vector minus a vector, terminal minus initial. Right? The midpoint of A B is half of, a vector plus b vector, in brackets. Simple, isn't it? And if P divides A B internally in the ratio m to n, written with a colon, so that A P is m and P B is n, then r, the orange arrow, is m times b vector, plus n times a vector, all over m plus n. Remember, the weight on each end is the length of the opposite part. Okay?

3:21Example 1: diagonals of a parallelogram bisect each other

Now, example one, and it is our puzzle from the start. Prove that the diagonals of a parallelogram bisect each other. Use the four moves. Pause the video and try it first. Done? Okay, let's check it together. Call the parallelogram ABCD, and take the vertex A as the origin, okay? Then name the two sides from that vertex: the vector A B is b vector, and the vector A D is d vector. These two sides are not parallel, so they are independent. Now, move three. What is the position vector of the fourth vertex, C? Yes, exactly, b vector plus d vector! Go from A to B, and then from B to C, and the side B C is the same vector as A D, because opposite sides of a parallelogram are equal and parallel. With me? Now take the diagonal A C, which starts right at the origin. So its midpoint is just half of c vector, which is half of, b vector plus d vector. Okay? And the midpoint of the other diagonal, B D, is half of, b vector plus d vector, straight from the midpoint formula. Right? Look, the two midpoints are the same point! So the diagonals meet at the middle of each one, which means they bisect each other. Puzzle solved, with no angles at all. See how clean that was?

4:42Example 2: ratios in which lines OC and BD cut each other

Now a harder one, JEE level. Read it with me. In triangle OAB, the vector O A is a vector, and the vector O B is b vector. Point C divides A B in the ratio one to two, and D is the midpoint of the side O A, right? The lines O C and B D meet at a point P, okay? Find the ratios O P to P C, and B P to P D, in which P cuts each line. Pause the video and try it first. Done? Alright, let's solve it. The origin is already O, and a vector and b vector are two sides of a triangle, so they are independent. Point C divides A B in the ratio one to two, so a vector gets the weight two and b vector gets the weight one: c vector is one third of, two a vector plus b vector. And the midpoint D is half of a vector. Got it? Now, P lies on the line O C, so it is some multiple of c vector: t times one third of, two a vector plus b vector, where t is some number. The small dot here simply means times. Okay? But P also lies on the line B D, right? For the points of B D, can I use the same letter t again? No, no, careful, that is the classic trap! P need not sit at the same fraction along both lines, so each line needs its own number, and on B D we call it s, a brand new letter. Start at B and move towards D: b vector plus s times the bracket, half of a vector, minus b vector. Clear? Both expressions are the same point P, so now comes move four: compare coefficients. Matching the a vector terms, two times t, over three, equals s over two. Matching the b vector terms, t over three equals one minus s. With me? From the first equation, s equals four times t, over three. Put this into the second, and five times t, over three, equals one. So t is three fifths, and s is four fifths. Right? Now, t equals three fifths means that, starting from O, P is three fifths of the way from O to C, along that line. So what is the ratio O P to P C? It is three to two, because P C is the remaining two fifths of the line O C, okay? And s equals four fifths means that P is four fifths of the way from B to D, so B P to P D is four to one. And P itself is one fifth of, two a vector plus b vector, and we are done.

7:18Quick revision: four moves and when to compare coefficients

So, quick revision. Choose an origin, name independent vectors, and write every point in terms of them. When two lines meet, give each line its own parameter, t and s. And compare coefficients, but only when the vectors are independent. Now go and try the practice problem on this page. You can do it. All the best!