Components, Position Vectors and the Section Formula

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. There is a point P on the segment from A to B, and the distance from A to P is three times the distance from P to B, okay? In the position vector of P, which end gets the bigger weight, A or B? Many students pick the wrong end. Today we crack it, along with components and position vectors, which come up often in JEE problems. So, let's begin.

0:26Components and position vectors

So, first the building blocks. Here, i hat, j hat and k hat, with little caps called hats, are the unit vectors along the x, y and z axes. Any vector a, with a small arrow on top, is a one times i hat, plus a two times j hat, plus a three times k hat, also written in brackets. The numbers with small subscripts are its components. Clear? Now take a point P with coordinates x, y, z, okay? The arrow over O P means the vector from the origin O to P, its position vector: x times i hat, plus y times j hat, plus z times k hat. Simple, right? Now tell me, if you know the three components, how will you find the length of the vector? Yes, exactly, Pythagoras in three dimensions! The vertical bars mean the modulus, the length. So mod a is the square root of a one squared, plus a two squared, plus a three squared; the small two on top means squared. Clear? And adding, subtracting, or multiplying by a number? The plus or minus sign means both cases, and lambda, the Greek letter, is any number. We simply work component by component. Got it?

1:37Direction cosines and the unit vector

Now, direction: divide each component by the modulus of a to get the direction cosines, cos alpha, cos beta and cos gamma, where these Greek letters are the angles the vector makes with the positive x, y and z axes. Of course, the vector must be non zero here, right? Now let me ask you this: if you square the three direction cosines and add them, what do you get? You get one, every single time! The little two on cos means squared, so cos squared alpha, plus cos squared beta, plus cos squared gamma, equals one. This works because we divided each component by exactly the length, okay? And here is the beauty. These three numbers are the components of the unit vector along a, which we call a hat: cos alpha i hat, plus cos beta j hat, plus cos gamma k hat. With me?

2:32From one point to another

Now, take two points, A and B, with position vectors a and b, okay? The vector from A to B, with an arrow over A B, is it a minus b, or b minus a? The answer is b minus a: the end point minus the starting point, terminal minus initial. Write a minus b and you get the right length, but the opposite direction. Clear? Next, parallel vectors: the two vertical lines mean is parallel to, and the double arrow means if and only if. So the components must be in the same ratio: a one over b one, equals a two over b two, equals a three over b three. Okay? And one small catch: a zero denominator means that numerator must also be zero. So if b three is zero, then a three is zero too. Right?

3:21The section formula

Now the main event, the section formula. In the figure, the arrows a and b go from the origin O to the points A and B, right? P divides A B internally in the ratio m to n, written with a colon, so the part A P is to the part P B as m is to n, okay? The orange arrow r is its position vector. Now look at the formula in the figure. The part m sits next to A, and yet m multiplies b, the far end. Can you see why? Here is the rule: the weight on each end is the length of the opposite part. So r equals m times b, plus n times a, all over m plus n. If m is bigger, P sits nearer to B, so B must pull harder. Got it? For external division, P is on A B extended, outside the segment. Both plus signs become minus: m times b, minus n times a, over m minus n. Okay? And the midpoint? Take m and n both equal to one, and you get half of a plus b. Nice and simple, isn't it?

4:26Centres and angle bisectors

Now the centres. Here a, b, c and d are the position vectors of the vertices, okay? The centroid of a triangle is one third of a plus b plus c, and of a tetrahedron, with four vertices, one fourth of a plus b plus c plus d, just averages. Easy, right? Now for the incentre, each vertex again gets a weight. Which weight do you think it is? Yes, the length of the opposite side, like the section formula! Plain letters without arrows are side lengths: small a is B C, small b is C A and small c is A B, okay? So the incentre is a times vector a, plus b times vector b, plus c times vector c, over a plus b plus c. Last, the angle between vectors a and b. The internal bisector is along a hat plus b hat, the external along a hat minus b hat. Notice the hats: we add the unit vectors, not a and b themselves, because the sum of two vectors bisects the angle between them only when the two have equal length, like the diagonal of a rhombus. With me?

5:27Example 1: vector PQ and its length, P(2,3,0) to Q(-1,-2,-4)

Now, example one, from the course. Find the vector from P, at two, three, zero, to Q, at minus one, minus two, minus four, and its length. Pause the video, and try it first. Done? Okay, let's check. The arrow over P Q means the vector from P to Q, so Q minus P: minus one minus two, minus two minus three, and minus four minus zero, okay? That gives minus three i hat, minus five j hat, minus four k hat. P minus Q would point the opposite way, so watch the order, right? For the length, the squares are nine, twenty five and sixteen, which add up to fifty, and the square root of fifty is five root two. Clear?

6:15Example 2: find P on AB with AP = 3PB, the weight trap

Now example two, and this is our puzzle from the start. P lies on the segment from A, at two, three, four, to B, at four, one, minus two, and A P is three times P B, okay? Where exactly is P? Pause the video, and try it first. Done? Okay, let's check. Since A P is three times P B, the ratio is three to one; the double arrow means implies. So, matching the figure, where the ratio A P to P B is m to n, m is three and n is one, clear? Now, three belongs to A P, so it goes with A, and P is one fourth of three A plus B, right? No, no, careful, that is exactly the trap from our puzzle! The long part is next to A, so P sits nearer to B, and B gets the weight three. So P is one fourth of three B plus A, okay? Three times B is twelve, three, minus six, and adding A gives twelve plus two, three plus three, and minus six plus four. With me? That is one fourth of fourteen, six, minus two. So P has x coordinate seven over two, y coordinate three over two, and z coordinate minus one over two. And yes, P is close to B, as the weights said!

7:29Quick revision: AB = b - a, unit vector, section formula

So, quick revision. The vector from A to B is b minus a, terminal minus initial. Divide vector a by its modulus to get the unit vector, a hat, whose components are the direction cosines. And in the section formula, the weight on each end is the length of the opposite part. Now go and try the practice problems on this page. All the best!