Areas, Collinearity, Distances and Moments

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Here is a line in space, and a point P not on it. How far is the point from the line? Not easy to picture in space, right? Today we will do it with just one cross product. The same tool gives areas of triangles, a test for collinear points, and the turning effect of a force. These ideas come up often in JEE papers. So, let's begin.

0:29Parallelogram: area from the cross product

First, a quick recall. In the figure, a vector and b vector start from the point O, at an angle theta; the small arrows on top mark vectors. The orange arrow is a cross b. It is perpendicular to both, the upside down T sign, with its direction from the right hand rule, from a to b. Its length, with two vertical bars for the modulus, is mod a times mod b times sine theta. Okay? Now look at the shaded parallelogram on a and b. The label inside says its area is mod a times mod b times sine theta, the same as the length of a cross b. But why is that true? Think of base times height, yes, exactly! The base is mod a, and the height, the dashed line marked h, is mod b times sine theta. Right? So the area is the modulus of a cross b, the length, not the vector itself. But what if you are given the two diagonals instead? Is the area still the modulus of their cross product? No, no, careful, let us check that. The diagonals are a plus b and a minus b. Here a cross a and b cross b are zero, and b cross a is minus a cross b, since the cross product is not commutative. So we get minus two times a cross b. Clear? Call the diagonals d one and d two, the small numbers below are just labels. Their cross product is twice as long as a cross b, so the area is half the modulus of d one cross d two. Forget this half, and you get double the answer. With me?

2:05Triangles and plane quadrilaterals

Now let us move on to triangles. Let me ask you this: a triangle is what part of a parallelogram? Exactly half, yes! One diagonal cuts a parallelogram into two equal triangles. So for the triangle ABC, take vector A B and vector A C, the long arrows on top. The area is half the modulus of their cross product. Got it? With position vectors a, b and c, vector A B is b minus a, and vector A C is c minus a. Expanding gives a cross b, plus b cross c, plus c cross a, in cyclic order. Half its modulus is the area. Okay? The same idea works for a plane quadrilateral ABCD, all four corners in one plane. Its area is half the modulus of vector A C cross vector B D, its two diagonals. Right?

2:56Three points on one line

Now here is a smart use of this. If three points lie on one straight line, what is the area of their triangle? Zero, yes, the triangle is completely flat! So the points are collinear exactly when vector A B cross vector A C is the zero vector, the zero with an arrow on top. The double headed arrow means if and only if. Clear? With position vectors, the test reads: a cross b, plus b cross c, plus c cross a, equals the zero vector. The zero vector, not the number zero, because a cross product is a vector. With me?

3:34Distance of a point from a line

Now, our puzzle from the start. Take a line through the point A along b vector, which is non zero; the equals sign with a slash means not equal to. And take a point P off the line. The cross product gives areas, so which parallelogram can help us? Yes, the one built on vector A P and b vector! Its area is the modulus of vector A P cross b. Okay? But the same area is also base times height. The base is mod b, and the height is exactly d, the distance of the point P from the line. Right? So, d equals the modulus of vector A P cross b, over the modulus of b. One cross product, one division, done! That is why b must be non zero, since we divide by its length. Got it?

4:25Moment of a force

One more use, from physics. When a force, F vector, acts at the point P, its moment about the point A, also called the torque, is M vector, equal to vector A P cross F vector. Now, could you write F cross vector A P instead? No, and this is a common trap. The cross product is not commutative, so swapping the order flips the sign, and you get minus M vector. Always the vector from A to P first, then the force. Okay?

4:54Example 1: area of triangle ABC from its three vertices

Now, example one, the easier one. Find the area of the triangle with vertices at the points A, one, one, two, then B, two, three, five, and C, one, five, five. Pause the video and try it first. Done? Okay, let's check. Vector A B is B minus A, which is one, two, three, and vector A C is C minus A, which is zero, four, three. The numbers in round brackets are the components. Right? Now the cross product, component by component. The first is two times three minus three times four, so six minus twelve gives minus six. In the same way, zero minus three gives minus three, and four minus zero gives four. With me? Its modulus is the root of thirty six plus nine plus sixteen, which is root sixty one. So the area is root sixty one, right? Oh no, careful, root sixty one is the area of the parallelogram! The triangle is half of it. So the area is root sixty one over two. Clear?

5:58Example 2: distance of a point from a line in space

Now example two, our puzzle from the start. Find the distance of the point P, three, minus one, two, from the line through the point A, one, zero, one, along two i hat plus j hat plus two k hat. The little caps mark the unit vectors along the axes. Pause the video and try it first. Done? Alright, let's solve it. Should we cross b vector with three, minus one, two, the position vector of P itself? No, no, that is the trap here. We need vector A P, which runs from the point A on the line to our point P, okay? So vector A P is P minus A, which is two, minus one, one, and b vector is two, one, two. Clear? Now let us cross them. The first component is minus one times two, then take away one times one, so minus two minus one gives minus three. In the same way, two minus four gives minus two, and two plus two gives four. Right? Its modulus is the root of nine plus four plus sixteen, which is root twenty nine. Got it? And mod b is the root of four plus one plus four, that is root nine, which is three. With me? So the distance is root twenty nine over three. Puzzle solved, in full three dimensions! See how fast that was?

7:14Quick revision: areas, collinearity, distance, moment

So, quick revision. For a parallelogram, use the modulus of a cross b from the sides, but half the modulus of d one cross d two from the diagonals. For a triangle, half the modulus of vector A B cross vector A C, and the zero vector means collinear. Distance from a line: the modulus of vector A P cross b, over mod b. And the moment is vector A P cross F vector. Now go and try the practice problems on this page. All the best!