Mixed Problems: Set 1 (Algebra and Dot Product)

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take two vectors, one of length three and one of length four. Add them, and the sum has length five. But three plus four is seven, not five! So where did the missing two go? Keep that question in mind. Today, in a mixed set on the dot product, one small habit, squaring the modulus, cracks this puzzle and many JEE Main questions like it. Problems of this type come up often, so let's begin.

0:33Tool 1: the dot product

So, tool number one, the dot product. The small arrow on top of a letter marks a vector, a quantity with a magnitude and a direction. The two vertical bars give the modulus, the length, so we read this as mod a. And theta, the Greek letter, is the angle between the two vectors, from zero up to pi, as the less than or equal signs show. So a dot b, with the small dot in the middle, equals mod a, times mod b, times cos theta. Okay? In components, multiply the matching parts and add: a dot b equals a one b one, plus a two b two, plus a three b three. The small numbers written low are subscripts, for the components along the three axes. Right? Rearrange the first line, and you get the angle: cos theta equals a dot b, over the modulus of a times the modulus of b. So an angle needs just one dot product and two lengths. Clear? Now let me ask you this. If theta is ninety degrees, and the small circle is the degree sign, what is a dot b? Yes, exactly, it is zero, because cos ninety degrees is zero! And it works both ways. The upside down T means perpendicular, and the double headed arrow means if and only if. But read the condition on screen: both vectors must be non zero, because the zero vector has no defined direction. With me?

1:57Tool 2: given a modulus? Square it

Now tool two, the most useful one in this set. Dot vector a with itself. The angle is zero and cos zero is one, so a dot a equals the modulus of a, squared; the small two on top means squared. So any length turns into a dot product the moment you square it. Keep the figure in view: the triangle law and the parallelogram law, which we use in a moment. Okay? Look at the triangle law on the left of the figure. The arrow from O to A is vector a, the arrow from A to B is vector b, and the orange arrow from O to B is a plus b. Now tell me honestly: is the length of a plus b just the two lengths added? No, no, careful, that is not true in general! So we square it. The modulus of a plus b, squared, is a plus b dotted with itself. Open the bracket, and you get the modulus of a squared, plus the modulus of b squared, plus two a dot b, because the two middle terms are equal. Right? For a minus b, the same steps put a minus sign on the middle term only. Now look at the parallelogram on the right: the orange diagonal is a plus b, and the green dashed one, from A to C, is b minus a, which has the same length as a minus b. Got it? So, back to our puzzle. Lengths do not simply add: the modulus of a plus b is less than or equal to the sum of the two lengths. This is the triangle inequality, because one side of a triangle is never longer than the other two together. Clear? And equality holds only for like parallel vectors, which point the same way. So three and four make seven only in that case; otherwise the sum is shorter. With me?

3:37Example 1: mod a = 3, mod b = 4, mod a+b = 5, find the angle

Now, example one, problem two of Set one, marked easy. The modulus of a is three, the modulus of b is four, and the modulus of a plus b is five. Find the angle between a and b. Yes, this is our puzzle from the start! Pause the video and try it first. Done? Okay, let's check. We are given a modulus, so we square it, with tool two. The modulus of a plus b, squared, equals the modulus of a squared, plus the modulus of b squared, plus two a dot b. Okay? Now put in the numbers. Five squared is twenty five, three squared is nine, and four squared is sixteen. So twenty five equals nine plus sixteen, plus two a dot b. Right? Look, nine plus sixteen is already twenty five. So what must a dot b be? Yes, exactly, it is zero! Twenty five minus twenty five leaves two a dot b equal to zero, so a dot b is zero. Clear? Now check the condition. The zero with an arrow on top is the zero vector, and the crossed equals sign means not equal. The lengths are three and four, so neither vector is the zero vector, and the double arrow means implies. So a dot b equal to zero really means perpendicular, and theta is ninety degrees. That is our puzzle solved: the triangle OAB, with sides three, four and five, is a right triangle, with the right angle at point A, between a and b placed head to tail. The figure is only a sketch, so that angle is not drawn as ninety degrees. See how fast that was?

5:17Example 2: unit vectors with a+b+c = 0, find mod a-b

Now a harder one, problem seven of the set, marked medium. Three unit vectors, a, b and c, add up to the zero vector. Unit vector means length one. Find the modulus of a minus b. Pause the video and try it first. Done? Okay, before we solve it, let me give you a quick test. Some students say that a and b both have length one, and one minus one is zero, so the answer is zero. Is that right? No, no, wait, that is the same mistake as our puzzle! Lengths do not subtract like numbers. So use the given equation. Move c to the other side: a plus b equals minus c. Now square the moduli. Minus c has the same length as c, so the modulus of a plus b, squared, equals the modulus of c, squared. Okay? Open the left side with tool two: one plus one, plus two a dot b, equals one, since all three are unit vectors. So two a dot b is minus one, and a dot b is minus one half. Right? Now the modulus of a minus b, squared, again with tool two. Careful with the signs here. What is minus two, times minus one half? Yes, it is plus one, because minus times minus is plus. So the modulus of a minus b, squared, is one plus one plus one, which is three. Got it? Now take the square root; the tick shaped sign on screen is the square root. So the modulus of a minus b is root three, and that is our answer. Let us check it another way. Cos theta is a dot b over the two lengths, one times one, so it is minus one half. So theta, the angle between a and b, is one hundred and twenty degrees. Clear? And here is a topper's check from the course: for unit vectors at an angle theta, the modulus of a minus b is two times sine of theta over two. Two times sine sixty degrees is two times root three over two, which is root three again. Perfect, both ways agree!

7:19Quick revision: perpendicular test, square the modulus

So, quick revision. For non zero vectors, a dot b equal to zero means perpendicular. Given a modulus? Square it: the modulus of a plus or minus b, squared, is the sum of the squared lengths, plus or minus two a dot b, as the plus or minus sign shows. And lengths do not simply add: the modulus of a plus b is at most mod a plus mod b. Now go and try the practice problems on this page. All the best!