Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Take two vectors, a and b, and make two new ones: a plus b, and a minus b. Suppose these two come out exactly the same length. Is that just luck, or does it tell us something about a and b? Keep your guess ready; Example two settles it. This is Part A of our vector algebra problem bank, at JEE Main level, built on familiar JEE question patterns. So, let's begin.

0:31Part A: a 30-minute test

So, here is what Part A holds. Ten solved problems, all at JEE Main level. Okay? And they are mixed: angles, perpendicular vectors, areas, volumes and work. You pick the tool yourself, just like in the exam. Right? The course plan: after Chapters one to five, attempt Part A as a thirty minute test. First, let us sharpen two tools and solve two problems together. With me?

0:59Tool 1: the dot product

Tool number one, from Chapter two: the dot product. Vector a dot b equals mod a times mod b times cos theta. The arrows on top mean vectors, the vertical bars mean the modulus, the length, and theta, the Greek letter, is the angle between them, from zero to pi; those signs mean less than or equal to. Okay? Now, suppose a question gives you a dot b, and both lengths. How do you get the angle? Yes, exactly, just divide! Cos theta equals a dot b, over mod a times mod b. Clear? In components, a dot b is a one times b one, plus a two times b two, plus a three times b three; the small numbers below are subscripts, for the x, y and z components. Now, if a dot b comes out zero, what does that tell you about the two vectors? They are perpendicular, yes! The upside down T means perpendicular, and the double headed arrow means if and only if. But the course says this for non-zero vectors only, so always check that word. Got it? And for the modulus of a sum or a difference, square it. Modulus of a plus or minus b, squared, equals mod a squared plus mod b squared, plus or minus two times a dot b. The small two means squared, and the sign with plus above minus means plus or minus. With me?

2:28Tool 2: the cross product

Tool number two, from Chapter three: the cross product. Vector a cross b equals mod a times mod b times sine theta, times n hat; the cross sign is the cross product, and n hat, with the little cap, is a unit vector. Okay? That unit vector n hat is perpendicular to both a and b; in the figure, a cross b is the orange arrow standing up from O, where both vectors start. For its direction, use the right-hand rule written on screen: curl the fingers of your right hand from a to b, and your thumb gives n hat. Clear? So, if we cross in the other order, b cross a, which way does the thumb point now? Downwards, yes, exactly, because the fingers now curl the other way! So b cross a equals minus a cross b. The cross product is not commutative, so the order matters. Right? Now let us look at its length. In the figure, the shaded parallelogram has base mod a, and the dashed height, h, is mod b times sine theta. So the modulus of a cross b is the area of the parallelogram. Got it? And a topper's tool from the course, Lagrange's identity: the modulus of a cross b, squared, equals mod a squared times mod b squared, minus the square of a dot b, in the round brackets. No angle needed at all, so keep it ready for Example one. With me?

3:53Example 1: modulus of a cross b when a dot b is 6 root 2

Now, Example one, from Part A, marked easy. Mod a is four, mod b is three, and a dot b is six root two; the tick-shaped sign is the square root. Find the modulus of a cross b. Pause the video and try it first. Done? Okay, let's check. Start with the angle, from Tool one. Cos theta is six root two, over four times three, which is twelve. Six root two over twelve gives one over root two. Okay? Now, cos theta is one over root two at forty five degrees, and also at three hundred and fifteen degrees, where sine theta is less than zero; the small circle means degrees, as the badge shows. So which angle is it? Forty five degrees, and only that! The angle between two vectors lies from zero to pi, and the double arrow means which gives. So sine theta is one over root two, and positive. Clear? Now we bring in Tool two. The modulus of a cross b is mod a times mod b times sine theta: twelve times sine forty five degrees. Twelve over root two gives six root two, the course answer. Right? Now a topper's check with Lagrange's identity. Mod a squared times mod b squared is sixteen times nine, which is one hundred and forty four. Six root two, squared, is seventy two. One hundred and forty four minus seventy two gives seventy two. Got it? So the modulus is root seventy two, which is six root two again. Same answer, and no angle needed. That is a neat check, isn't it?

5:36Example 2: equal moduli of a + b and a - b, so x = -1

Now, remember our puzzle? Here it is with real vectors, a medium one. Vector a is x times i hat, plus two j hat, plus k hat, and vector b is i hat, minus j hat, plus three k hat; the little caps mark the unit vectors along the axes. The modulus of a plus b equals the modulus of a minus b, and we have to find the value of x, okay? Pause the video and try it first. Done? Okay, let's check. Some students see a on both sides and simply cancel it. Then b equals minus b, so b must be the zero vector. Is that allowed? No, no, wait, that is the trap! Equal lengths do not mean equal vectors. So use Tool one: square both moduli and subtract. Mod a squared and mod b squared cancel, leaving four times a dot b. Okay? The moduli are equal, so this difference is zero, and a dot b is zero. Both vectors are clearly non-zero, so a is perpendicular to b, and that answers our puzzle. Right? Now look at the right half of the figure, the parallelogram law; we can leave the triangle law on the left aside. The orange diagonal is a plus b, and the dashed green one is b minus a, which has the same length as a minus b. Equal diagonals happen exactly when the parallelogram is a rectangle. Clear? Now let us put in the numbers. In components, a dot b is x times one, plus two times minus one, plus one times three. That is x minus two plus three, and it must be zero. Got it? So x plus one is zero, which gives x equal to minus one. That matches the course answer, very good! Let us check. With x equal to minus one, a plus b has components zero, one, four, written in round brackets, and a minus b has components minus two, three, minus two. Both have modulus root seventeen, just as the question says. Beautiful, isn't it?

7:42Quick revision: angle, Lagrange identity, perpendicular test

So, quick revision. Cos theta is a dot b over mod a times mod b, with theta from zero to pi, so sine theta is greater than or equal to zero. The modulus of a cross b is mod a times mod b times sine theta, or use Lagrange's identity. For non-zero vectors, a dot b equal to zero means perpendicular, and for the modulus of a sum, square it. Now set a thirty minute clock, attempt Part A, then try the practice problem on this page, and all the best!