Definition, Properties and Computation

Transcript

0:00Introduction

Hello everyone! With numbers, two times three equals three times two, right? And if a number times itself is zero, the number is zero. Vectors can break both rules! Swap the order, and the answer flips. Cross a non-zero vector with itself, and you get the zero vector. How is that possible? It is the cross product, which comes up often in JEE questions. Let's begin.

0:28The vector (cross) product

So, take vectors a and b at an angle theta, the Greek letter; the arrow on top marks a vector. Their cross product, read as a cross b, equals mod a, times mod b, times sine theta, times n hat; the cap on n, called a hat, marks a unit vector. The vertical bars mean the modulus, the length. Okay? So n hat is a unit vector, of length one. The upside down T means perpendicular, so n hat is perpendicular to both a and b, and the cross product is a vector! But up and down are both perpendicular to this plane. Which one do we take? We use the right-hand rule. Curl the fingers of your right hand from a towards b, through the angle theta, and your thumb gives n hat. In the figure, that is upwards. Clear? Now, the length of a cross b is mod a times mod b times sine theta. In the figure, that is the base, mod a, times the dashed height, h, which is mod b sine theta. So it is the area of the parallelogram. Got it?

1:34Properties of the cross product

Now, the big question. What about b cross a? Curl your fingers from b towards a this time. Which way does your thumb point now? It points the opposite way, yes, exactly! Same length, opposite direction, so b cross a equals minus a cross b. It is anticommutative, so the order matters. That is the first part of our puzzle. Right? The bracket rules still work. Vector a cross, b plus c, equals a cross b plus a cross c, and a number lambda, the Greek letter, comes straight out. Now, what is a cross a, a vector crossed with itself? It is the zero vector, the zero with an arrow, since sine zero is zero, and that is puzzle number two solved! For non-zero vectors, a cross b is zero if and only if a is parallel to b. The two vertical lines mean parallel; the double headed arrow means if and only if. With me?

2:32Unit vectors and the determinant

Now i hat, j hat and k hat, along the axes. i hat cross j hat is k hat, j hat cross k hat is i hat, and k hat cross i hat is j hat, a cycle. So what is j hat cross i hat? It is minus k hat, that is correct, because we went against the cycle, and so on for the other pairs. And i hat cross i hat is the zero vector. Okay? Now the working formula. Vector a has components a one, a two, a three, and b has b one, b two, b three; the small numbers written low are subscripts. Then a cross b is this determinant, the grid between vertical lines. Clear? Expanding gives three components, in brackets. First, a two b three minus a three b two; then a three b one minus a one b three; then a one b two minus a two b one. See the trick? The subscripts go two three, three one, one two, the same cycle! Got it?

3:34Length, sine and perpendicular unit vectors

Next, Lagrange's identity. Mod of a cross b, squared, equals mod a squared times mod b squared, minus, a dot b, squared; that is the dot product, written with the small dot. So the cross product and the dot product are linked. Right? So sine theta equals mod of a cross b, over mod a times mod b, for non-zero a and b. Now tell me, how many unit vectors are perpendicular to both a and b? There are two, yes, exactly, one on each side! Divide a cross b by its length and put plus or minus in front; the slashed equals sign says a cross b must not be zero. Okay?

4:15Trap: no cancelling, and brackets matter

Now, a classic JEE trap. Suppose a cross b equals a cross c, with a not the zero vector. Just cancel vector a from both sides, and b equals c, right? No, no, wait, that is not allowed! Bring everything to one side: a cross, b minus c, is zero. So a is parallel to b minus c, and b equals c plus t times a, where t is some number. Got it? And one more trap: the cross product is not associative. Vector a cross b, then cross c, is in general different from a cross, b cross c. So keep the brackets. Clear?

4:54Example 1: cross product of 2i + j - k and i - j + 2k

Okay, now a course example. Find two i hat plus j hat minus k hat, cross, i hat minus j hat plus two k hat. Pause the video and try it first. Done? Okay, let's check. Call them a and b. So a one, a two, a three are two, one, minus one, and b one, b two, b three are one, minus one, two. Okay? First, the i hat part: one times two, minus, minus one times minus one. So is it two plus one, or two minus one? It is two minus one, which is one, from a two b three minus a three b two. Minus one times minus one is plus one, and we subtract it. Right? Second, a three b one minus a one b three. Minus one times one, minus two times two, so minus one minus four, which gives minus five. Clear? Third, a one b two minus a two b one. Two times minus one, minus one times one, so minus two minus one, which gives minus three. Okay? So vector a cross b equals i hat, minus five j hat, minus three k hat. With me? Now a free check: a cross b is perpendicular to a, so their dot product must be zero. Two minus five plus three gives zero, perfect! Got it?

6:13Example 2: vector of magnitude 9 perpendicular to both

Now a harder one. Find a vector of magnitude nine, perpendicular to both four i hat minus j hat plus three k hat, and minus two i hat plus j hat minus two k hat. Find a unit vector first. Pause the video and try it. Done? Okay, let's check. Call them a and b. First, minus one times minus two, minus three times one, so two minus three. Second, three times minus two, minus four times minus two, so minus six plus eight. Third, four times one, minus, minus one times minus two, so four minus two. So a cross b is minus one, two, two. Okay? Now, nine times a cross b is also perpendicular to both. So that is our answer, right? Oh no, careful, that is exactly the trap! Vector a cross b is not a unit vector. Its length is the square root, the tick shaped sign, of one plus four plus four, which is three. So nine times a cross b has length twenty seven. Clear? So first make a unit vector: divide by three. We get plus or minus one third, times minus i hat plus two j hat plus two k hat. Both signs, remember? Now multiply by nine: nine over three gives three. So the answer is plus or minus three, times minus i hat plus two j hat plus two k hat. Neat, right?

7:41Quick revision: formula, order, perpendicular unit vectors

So, quick revision. Vector a cross b is mod a times mod b times sine theta, times n hat, a vector perpendicular to both by the right-hand rule. Order matters: b cross a is minus a cross b, and a cross a is zero. Unit vectors perpendicular to both are plus or minus a cross b over its length. And never cancel vector a from a cross product! Now try the practice problems on this page. All the best!