Mixed Problems: Set 2 (Cross Product and Areas)

Transcript

0:00Introduction

Hello everyone! Let us start with a quick puzzle. Three points in space: the origin, the point one, two, three, and the point three, two, one. What is the area of the triangle they make? We are given no angles and no side lengths. Hmm, tricky, right? By the end of this video, you will do it in about thirty seconds, with just one cross product. Problems like this come up often in JEE Main. So, let's begin.

0:30Tool 1: the cross product

First, our main tool. In the figure, a vector and b vector start from the point O, with the angle theta between them; the small arrows on top mark vectors. The orange arrow is a cross b; the cross sign between them, like a times sign, means the cross product. It equals mod a times mod b times sine theta, times n hat. The vertical bars mean the modulus, the length, and the cap on n makes it a unit vector. Okay? Now, the direction of n hat. It is perpendicular to both a and b; the upside down T sign means perpendicular. Curl the fingers of your right hand from a to b, and your thumb gives n hat. That is the right hand rule. With me? Now tell me this. If a cross b points straight up, like the orange arrow, which way does b cross a point? Straight down, yes, exactly! Curl your fingers from b to a, and the thumb flips over. So b cross a equals minus a cross b; the cross product is not commutative. And a cross a is the zero vector, the zero with an arrow on top, since sine zero is zero. Keep this pair ready, we need it in example two. Clear? And now in components. Write a as a one, a two, a three, and b as b one, b two, b three; the small numbers below are just labels. In round brackets, a cross b has three components. The first is a two times b three, minus a three times b two. The second and third follow the same cyclic pattern, as you see on screen. Got it?

2:03Tool 2: areas from one cross product

Now look at the shaded parallelogram in the figure, built on a and b. Base times height: what is its area? The base is mod a, and the height, the dashed line marked h, is mod b times sine theta. So the area is the modulus of a cross b, the length of the orange arrow, not the vector itself. With me? Now, every triangle is exactly half of a parallelogram, cut along its diagonal. So for the triangle ABC, take vector A B and vector A C, with the long arrows on top. The area is half the modulus of their cross product. Okay? And sometimes a question gives the two diagonals, d one and d two, instead of the sides. Then the area is half the modulus of d one cross d two. Forget that half, and your answer doubles. Right?

2:57Tool 3: two signs to watch

Next, Lagrange's identity. The modulus of a cross b, squared, equals mod a squared times mod b squared, minus a dot b, squared. The small two on top means squared, and the dot is the dot product. It is true because the cross product brings sine theta, the dot product brings cos theta, and sine squared plus cos squared is one. Clear? Now suppose a question gives you only three lengths: mod a, mod b, and the modulus of a cross b. How many values can a dot b take then, one or two? In general, two values, yes, exactly! The identity gives only the square of a dot b. Theta, and one hundred and eighty degrees minus theta, have the same sine but opposite cosines. So write plus or minus, the sign on screen; writing only the positive value is a common slip. Got it? The same two signs appear here. The unit vectors perpendicular to both a and b are plus or minus a cross b, over the modulus of a cross b. This needs a cross b to be non zero, so a and b must be non zero and not parallel. So there are two answers, pointing opposite ways. Okay?

4:08Example 1: area of triangle O, (1,2,3), (3,2,1)

Now example one, the easier one, and it is our puzzle from the start. Find the area of the triangle with vertices O, the origin, then one, two, three, and three, two, one. Pause the video and try it first. Done? Okay, let's check. One corner is the origin O itself. So which two vectors give the sides from the origin? Yes, exactly, just the position vectors of the other two corners: one, two, three, and three, two, one. Because O is the origin, there is nothing to subtract. The numbers in round brackets are the components. Right? Now tool one, component by component. The first is two times one, minus three times two, so two minus six. The second is three times three, minus one times one, so nine minus one. And the third is one times two, minus two times three, so two minus six. With me? So the cross product is the vector minus four, eight, minus four. Okay? Its length is the root of sixteen plus sixty four plus sixteen, that is, root ninety six; the tick shaped sign is the square root. And root ninety six is four root six. So the area of our triangle is four root six, right? Oh no, careful, four root six is the area of the parallelogram! The triangle is only half of it. Half of four root six is two root six. Puzzle solved, with one cross product. See how fast that was?

5:42Example 2: parallelogram with sides a + 2b and 2a + b

Now example two, a little harder. Here, a vector and b vector are unit vectors, at an angle of thirty degrees; the small circle means degrees. Find the area of the parallelogram with sides a plus two b, and two a plus b. Pause the video and try it first. Done? Okay, let's check. The area of a parallelogram is the modulus of the cross product of its sides. So first we find a plus two b, cross two a plus b. Okay? Expand it just like ordinary brackets, but keep the order in every product. We get two times a cross a, plus a cross b, plus four times b cross a, plus two times b cross b. Right? Now, from tool one, a cross a and b cross b are both the zero vector. So only a cross b, plus four times b cross a, is left. Got it? Now, many students simply add here: one plus four, so five times a cross b. Is that correct? No, no, wait, that is the trap here! Remember, b cross a is minus a cross b. So we get a cross b, minus four times a cross b, which is minus three times a cross b. The five would have given you five over two, a wrong answer. Clear? The area is the modulus, so the minus sign goes away. It is three times mod a times mod b times sine thirty degrees. Both lengths are one, and sine thirty degrees is one half. So the area is three times one half, which is three over two. With me?

7:17Quick revision: order, half for triangles, plus-minus signs

So, quick revision. First, b cross a is minus a cross b, and a cross a is the zero vector, so always keep the order. Second, the area from the sides is the modulus of their cross product; for a triangle, or from the diagonals, take half. And third, when a dot b comes from lengths alone, write plus or minus; and the unit vectors perpendicular to both a and b always come as a plus or minus pair. You spotted two real traps today, so well done. Now go and try the practice problems on this page. All the best!