Definition, Properties and Angles

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. If two ordinary numbers multiply to give zero, then one of them must be zero, right? Today we will multiply two vectors, neither of them zero, and get exactly zero! And the answer is not even a vector, just a plain number. How is that possible? The secret is the dot product. It comes up often in JEE vector questions. So, let's begin.

0:28The dot product

So, take two vectors, a and b; the small arrow on top marks a vector. Their dot product, with a small dot in the middle, is read as a dot b, okay? It equals mod a, times mod b, times cos theta. The vertical bars give the modulus, the length of the vector, and theta, the Greek letter, is the angle between them. Clear? Theta runs from zero up to pi, which is one hundred and eighty degrees, and the small circle is the degree sign. The little signs mean less than or equal to. Now tell me, is a dot b a vector, or just a number? Just a number, yes, exactly! Mod a, mod b and cos theta are all numbers, so their product is a scalar. That is why we call it the scalar product. Right?

1:18Dot product in components

Now, in components. Vector a is a one i hat, plus a two j hat, plus a three k hat, and vector b likewise, with b one, b two, b three. The small numbers written low are subscripts, and the caps, called hats, mark the unit vectors along the x, y and z axes. Now, what is i hat dot i hat? Yes, the answer is one! Both lengths are one and the angle is zero, so we get one times one times cos zero. Same for j hat and k hat. Okay? Next, i hat dot j hat. These two are at right angles, so cos theta is zero, and so is the dot product. The same for the other two pairs. With me? So when we multiply out, only the matching products survive, and a dot b equals a one b one, plus a two b two, plus a three b three, that's it! Got it?

2:18Properties of the dot product

Alright, now the rules. Order does not matter: a dot b equals b dot a. And a scalar lambda, the Greek letter on screen, comes straight out: lambda a in brackets, dot b, equals lambda times a dot b. Right? Next, a dot, b plus c in brackets, equals a dot b, plus a dot c. Now, what is a dot a, a vector dotted with itself? Mod a squared, yes, exactly! The angle is zero, so we get mod a times mod a. Remember this line. It turns any length into a dot product.

2:57The angle between two vectors

Clear? Now the most useful part. Divide by mod a times mod b, and cos theta equals a dot b, over mod a times mod b. We divide by the lengths, so both vectors must be non-zero. If a dot b is greater than zero, what kind of angle is theta? It is acute, that is correct! But careful: theta can also be zero, when the vectors are like parallel, that is, pointing the same way, so a strictly acute angle needs them not like parallel. Right? And if a dot b is less than zero, cos theta is negative, so the angle is obtuse. The same care here: theta can be one hundred and eighty degrees, when the vectors are unlike parallel, that is, pointing opposite ways, so a strictly obtuse angle needs them not unlike parallel. Okay? For non-zero vectors, vector a is perpendicular to vector b if and only if a dot b is zero. The upside down T means perpendicular, and the double headed arrow means if and only if. That solves our puzzle: two non-zero vectors at right angles have a zero dot product. With me?

4:06Topper's trick: square the modulus

Now a topper's trick. When you see the modulus of a sum, square it, and write it as a dot product with itself. So mod of a plus or minus b, squared, equals mod a squared, plus mod b squared, plus or minus two times a dot b; the sign with plus over minus means plus or minus. Take the top signs together, or the bottom signs together. Okay? Similarly, vector a plus b, dot, a minus b, equals mod a squared, minus mod b squared. The middle terms cancel, because a dot b equals b dot a. Right? And for three vectors, the same idea gives the three squares, plus two times the sum of the three dot products. Clear?

4:49Trap: you cannot cancel a vector

Now, a classic JEE trap. Suppose a dot b equals a dot c. Just cancel vector a from both sides, and b equals c, right? No, not always, so be careful here! We cannot cancel a vector. Bring everything to one side: a dot, b minus c, equals zero. That only says vector a is perpendicular to b minus c, so b and c can still be different. Got it? Here is one more trap: every dot product is just a number. So a dot b, times c, is parallel to c, while vector a, times b dot c, is parallel to vector a, so in general they differ. Clear?

5:31Example 1: angle between i + 2j + 3k and 3i + 2j + k

Okay, now a course example. Find the angle between i hat plus two j hat plus three k hat, and three i hat plus two j hat plus k hat. Pause the video and try it first. Done? Okay, let's check. Call the first vector a and the second vector b. Multiply the matching components: one times three, two times two, and three times one. Three plus four plus three gives ten. Okay? Now the lengths. The tick shaped sign is the square root. Mod a is the root of one plus four plus nine, which is root fourteen, and mod b is the root of nine plus four plus one, also root fourteen. Right? So cos theta equals ten, over root fourteen times root fourteen. That is ten over fourteen, which is five over seven. Now tell me: is this angle acute or obtuse? It is acute, yes, because the dot product, ten, is positive. So theta is cos inverse of five over seven. The minus one up there is not a power; it means cos inverse. Got it?

6:42Example 2: a + b + c = 0, find the angle between a and b

Now a harder one, JEE style. Vector a plus vector b plus vector c equals the zero vector, the zero with an arrow on top. Mod a is three, mod b is five, and mod c is root nineteen. Find the angle between a and b. Here is a hint: square a modulus. Pause the video and try it first. Done? Okay, let's check. Move c across: c equals minus, a plus b in brackets. The minus sign flips the direction, not the length, so mod c squared equals mod of a plus b, squared. Okay? Now expand it. Mod c squared is nineteen, mod a squared is nine, and mod b squared is twenty five. So nineteen equals nine plus twenty five, plus two times a dot b. Right? Nine plus twenty five is thirty four, so two times a dot b equals nineteen minus thirty four, which is minus fifteen. So a dot b is minus fifteen over two. Clear? Now the definition. Mod a times mod b is three times five, which is fifteen. So fifteen cos theta equals minus fifteen over two, and cos theta is minus one half. Which angle between zero and one hundred and eighty degrees has cosine minus one half? One hundred and twenty degrees, exactly! See, no components at all, just one squared modulus. Got it?

8:14Quick revision: dot product, perpendicular test, modulus

So, quick revision. The dot product is mod a times mod b times cos theta, or the sum of the products of matching components. For non-zero vectors, it is zero exactly when they are perpendicular. And square a modulus using a dot a equals mod a squared, but never cancel a vector. Now go and try the practice problems on this page. All the best!