Overview

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle before we start. With numbers, if r times a equals b times a, and the number a is not zero, we cancel it, and r equals b, right? Now let them be vectors, with the cross product: r cross a equals b cross a. Must r equal b? Hold on to your answer for now. This chapter is a set of mixed vector problems in JEE Main style, and vector questions come up often in JEE Main. So let us see how to use it, and solve two of its problems together. Let's begin!

0:35How to use this chapter

So, let's open this chapter. The course names four kinds of vector questions that come up often in JEE Main. Type one: an angle between two vectors, or a projection of one vector on another. Okay? Type two: an area, of a triangle or a parallelogram. Right? Type three: a vector equation, like r cross a equals b cross a. The small arrow on top marks a vector, and the cross sign is the cross product. Yes, that is our puzzle, and we settle it very soon. With me? And type four: a triple product identity, from the scalar or the vector triple product. Clear? Now, about the real questions. The actual JEE Main questions from twenty twenty one to twenty twenty six are at the end of Chapters one to four; the short dash means to. Got it? And here, in this chapter, you get three sets of original problems in the same style. Each one should take you one to two minutes. So solve them with a clock running. Okay?

1:37The type 3 trap: can we cancel a vector?

So, back to our puzzle. Vector r cross a equals b cross a, where a is not the zero vector; the slashed equals sign means not equal, and the zero with an arrow is the zero vector. Now tell me, can we cancel a, and say r equals b? No, no, wait, we cannot cancel a vector! Instead, bring everything to one side. Then, r minus b in brackets, cross a, is the zero vector. Right? Now look at the figure. Vectors a and b start from the point O, with the angle theta, the Greek letter, between them. The orange arrow is a cross b, and the upside down T means perpendicular: a cross b is perpendicular to both a and b, with its direction from the right-hand rule, curling from a towards b, as the little arrow in the note shows. Its length, mod a times mod b times sine theta, is the shaded area, the base mod a times the dashed height h; the vertical bars mean the modulus. Clear? Now recall that for non-zero vectors, a cross product is zero only when sine theta is zero, that is, when the vectors are parallel. So r minus b is either zero or parallel to a; the two vertical lines mean parallel. Either way, r equals b plus t times a, for some number t. Not one answer, but a whole line of answers! So r need not equal b. With me? Then one unknown number, t, is left. To fix it, the question gives one more condition, a scalar one, like a dot product equal to a number. Remember this move; we use it in Example two. Got it?

3:15Example 1: |a| = 3, |b| = 4, |a + b| = 5, find the angle

Now, Example one, an easy one from Set one. Mod a is three, mod b is four, and mod of a plus b is five. Find the angle between a and b. Aim for one to two minutes. Pause the video and try it first. Done? Okay, let's check. But be careful here, mod of a plus b is not mod a plus mod b, because three plus four gives seven, and the question says five. Here is the trick: when a question gives 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. The small two on top means squared, and the dot in the middle is the dot product. Okay? Now put in the numbers. Twenty five equals nine plus sixteen, plus two times a dot b. So what is a dot b? Yes, exactly, it is zero! Nine plus sixteen is already twenty five, so two times a dot b is zero. The double arrow means implies, so a dot b equals zero. Right? Now the angle. Cos theta equals a dot b, over mod a times mod b, and that is zero. So theta is ninety degrees; the small circle means degrees. For non-zero vectors, a zero dot product means perpendicular. And see, three, four, five is a Pythagorean triple, which is why the angle came out as a right angle. Got it?

4:43Example 2: solve r x a = b x a with r . (i + 2j) = 0

Now Example two, a medium one from Set two, and it is our puzzle. Vector a is i hat minus j hat, and vector b is j hat plus k hat; the hats mark the unit vectors along the axes. Find r, such that r cross a equals b cross a, and r dot, i hat plus two j hat, equals zero. Pause the video and try it first. Done? Okay, let's check. First, the tempting shortcut says that r is just b, so let us test it. Vector b dot, i hat plus two j hat, is zero plus two plus zero, which is two. But we need zero, so r equals b is wrong. So our puzzle is settled: here, r is not equal to b. Clear? Now the proper way, by our move: r equals b plus t times a. In components, in round brackets, b is zero, one, one, and a is one, minus one, zero. So r is t, then one minus t, then one. Okay? Now the dot condition. Dot r with i hat plus two j hat. What value of t makes it zero? Let's work it out. It is t, plus two times, one minus t, which simplifies to two minus t. Two minus t equals zero, so t equals two. Right? Put t equal to two. Then r is two, minus one, one. So r equals two i hat minus j hat plus k hat. With me? And always check. Vector r minus b is two, minus two, zero, which is two times a. So r minus b, cross a, is two times a cross a, the zero vector. Both conditions hold, and the answer is right. Got it?

6:28Revision: four types, square |a + b|, never cancel a vector

So, quick revision. Four common types: an angle or a projection, an area, a vector equation, and a triple product identity. Given the modulus of a sum? Square it, and use a dot b. And never cancel a vector: r cross a equals b cross a, with a non-zero, gives r equals b plus t times a. Now open Set one, give each problem one to two minutes, and start solving. All the best!