Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. For two vectors, suppose a dot b is zero. Then they are perpendicular, right? And suppose a cross b is the zero vector too. Then they are parallel. Perpendicular and parallel, at the same time? Hold on to your answer for now. This chapter is the problem bank for the whole of vector algebra, and that puzzle is one of its concept checks. So let us see how to use this chapter, and solve two of its problems together. Let's begin!

0:33How to use this chapter

So, let's open this chapter. Part A has ten solved problems, at JEE Main level. Okay? And they are mixed: dot products, cross products and triple products, all in one set. So you choose the tool yourself. Right? Now, the course says: attempt Part A as a thirty minute test, after you finish Chapters one to five. Clear? So, a quick question for you. Ten problems in thirty minutes. How much time can you give each problem? Thirty over ten gives three. So three minutes a problem, with a clock running. Got it? Then Part B has eight problems at JEE Advanced level, each combining several ideas. With me? And Part C has six concept checks. They are statements often tested as true or false. Okay?

1:24Part C: true or false, with a reason

Now, two statements from Part C, and we answer each one with a reason. The first one is our puzzle. Vector a dot b is zero, and a cross b is the zero vector. The small arrow on top marks a vector, the dot is the dot product, the cross sign is the cross product, and the zero with an arrow is the zero vector. So, what can be said about a and b? Let us write both in terms of lengths. Vector a dot b is mod a times mod b times cos theta; the vertical bars mean the modulus, the length, and theta, the Greek letter, is the angle between them. And the length of a cross b is mod a times mod b times sine theta. Both are zero. Right? But cos squared theta plus sine squared theta is always one; the small two on top means squared. So cos theta and sine theta are never both zero. Then mod a times mod b must be zero. Clear? So at least one of the vectors is the zero vector, and our puzzle is solved! For two non-zero vectors, this can never happen. That is why the rules say non-zero: for non-zero vectors, a dot b equal to zero means perpendicular, and a cross b equal to the zero vector means parallel. Always check that word. Got it? Now the second statement. Is the cross product associative? That means, can we move the brackets freely, the way we do with numbers? Let us test it with i hat, j hat and k hat, the unit vectors along the axes; the little cap on top is the hat. First the bracket: i hat cross j hat is k hat. Then k hat cross j hat is minus i hat. With me? Now move the brackets. Here j hat cross j hat is the zero vector, since any vector crossed with itself gives zero. So i hat cross the zero vector is the zero vector. Minus i hat on one side, the zero vector on the other, so the answer is no: the cross product is not associative. In Part C, one such counterexample is enough to call a statement false. Okay?

3:36Example 1: angle between a and b when a + b + c = 0

Now, Example one, from Part A, marked medium. Vector a plus b plus c is the zero vector, and mod a is three, mod b is five, and mod c is seven. Find the angle between a and b. Remember, three minutes. Pause the video and try it first. Done? Okay, let's check. Since a plus b plus c is the zero vector, c equals minus, a plus b in brackets. The minus sign flips the direction, but not the length. So mod c squared equals mod of a plus b, squared. Right? Now the trick for the modulus of a sum: square it. Mod of a plus b, squared, equals mod a squared plus mod b squared, plus two times a dot b. Clear? Put in the numbers: forty nine equals nine plus twenty five, plus two times a dot b. So what is a dot b? Nine plus twenty five is thirty four, and forty nine minus thirty four is fifteen. So two times a dot b is fifteen, which gives a dot b equal to fifteen over two; the double arrow means which gives. Got it? Now bring in the angle. Vector a dot b is mod a times mod b times cos theta, and three times five is fifteen. So fifteen cos theta equals fifteen over two. With me? So cos theta is one half, and theta is sixty degrees; the small circle means degrees. That is the course answer. Okay? But wait, there is one more thing to check. Placed head to tail, the vectors a, b and c form a triangle, with sides three, five and seven. Is the angle of that triangle, between the sides three and five, also sixty degrees? No, it is not, and this is the trap! Head to tail, the triangle's angle between the sides three and five is one hundred and eighty minus sixty, which is one hundred and twenty degrees. But the angle between two vectors is measured tail to tail, so the answer stays sixty degrees. Right?

5:44Example 2: modulus of a + b + c is root 14, not 6

Now a harder one, from Part B, marked hard. Mod a is one, mod b is two, and mod c is three. And each vector is perpendicular to the sum of the other two. Find the modulus of a plus b plus c. Pause the video and try it first. Done? Okay, let's check. Some students simply add the lengths: one plus two plus three is six. So the answer is six, right? No, no, careful, that is the trap! The modulus of a sum is not the sum of the moduli. So use the condition, where perpendicular means the dot product is zero. Vector a is perpendicular to b plus c, so a dot, b plus c, is zero, and the same for b and for c. Okay? Now the Part B idea: add all three equations. What do you get? Open the brackets, and each dot product appears twice, since a dot b equals b dot a. So two times, a dot b plus b dot c plus c dot a, equals zero. Clear? Now the same trick as Example one: square the modulus of the sum. For three vectors, mod of a plus b plus c, squared, equals the three squares added, plus two times that same bracket. Right? And that bracket is zero! So we get one plus four plus nine, plus zero, which is fourteen. Got it? So the modulus is root fourteen; the tick-shaped sign is the square root. That is the course answer, and it is not six. Two ideas in one question: that is what Part B is like. With me?

7:26Quick revision: dot and cross both zero, square the sum

So, quick revision. Part A is a thirty minute test, Part B is at JEE Advanced level, and Part C tests statements as true or false. If a dot b is zero and a cross b is the zero vector, then at least one of the vectors is the zero vector. For the modulus of a sum, square it, and measure the angle between two vectors tail to tail. Now open Part A, set a thirty minute clock, and start solving, then try the practice problems on this page. You can do this. All the best!