Mixed Problems: Set 3 (Triple Products and Vector Equations)

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take three arrows from one corner, with no two of them parallel. Usually they build a slanted box with a real volume. But tilt one arrow just right, and the box goes completely flat. So how do you catch that exact tilt, with one single number? Today, in Set three of our mixed problems, we use the scalar triple product and the vector triple product. Questions on these come up often in JEE Main, so let's begin.

0:33Tool 1: the scalar triple product

So, tool one, the scalar triple product. The small arrow on top of each letter marks a vector, with a magnitude and a direction. Vectors a, b and c in square brackets mean a dot, b cross c. The round brackets say: find b cross c first, with the cross sign, then take its dot product with vector a, the small dot. So the result is just a number. Okay? In components, it is this determinant; the two vertical lines around the array mean determinant. The components of a, b and c fill the three rows, and the small numbers written low are subscripts, for the parts along the three axes. Right? Now look at the figure. Vectors b and c, from the corner O, make the shaded base, and b cross c, the orange arrow, is perpendicular to it. The dashed line h, the height, is the projection of vector a on b cross c. The note on the right says volume is base times height. But let me ask you this: can this triple product be negative? Yes, it certainly can be negative! It is positive when vectors a, b and c form a right handed system, and swapping two of them flips the sign. Since a volume cannot be negative, the volume of the parallelepiped is the modulus of a b c in square brackets; the outer vertical bars are the modulus. Clear? Now, back to our puzzle. When does the box go flat? When all three vectors lie in one plane, the height is zero, so the volume is zero. So vectors a, b and c are coplanar if and only if a b c in square brackets is zero; the double headed arrow means if and only if. With me?

2:17Tool 2: the vector triple product

Now tool two, the vector triple product: vector a, cross, b cross c in brackets. Two cross products in a row look scary, right? But it opens in one line, as a dot c, times vector b, minus a dot b, times vector c, okay? To remember the order, the course gives a rule: outer dot far, times near, minus outer dot near, times far. Here vector a is the outer one; inside the brackets, b is near and c is far. So outer dot far is a dot c, and it goes with b, the near vector. Got it? Now look at the answer again: it is built only from b and c. So tell me, in which plane does it lie? In the plane of b and c, yes, exactly! And since it is a cross product with vector a, it is also perpendicular to vector a. Clear? Now, what if the bracket moves to the left, so we take a cross b first, then cross c? Same answer, or not? Vector a cross b, then cross c, equals a dot c, times b, minus b dot c, times vector a, as you see on screen. Look, this one is built from b and a, not from b and c. Right? So the answer really does change! In general these two are not equal; the crossed equals sign means not equal. This is a common trap, so check where the bracket sits before you expand. With me?

3:45Example 1: find lambda so the three vectors are coplanar

Now, example one, problem twenty two, marked easy. Find lambda, the Greek letter, so that these three vectors are coplanar. The little caps are hats; i hat, j hat and k hat are the unit vectors along the three axes. The vectors are i hat plus two j hat plus lambda k hat, then two i hat plus j hat plus k hat, and i hat minus j hat plus two k hat. See, this is our flat box puzzle! Pause the video and try it first. Done? Okay, let's check. Coplanar means the scalar triple product is zero. So the components become the rows of a determinant. The first row is one, two, lambda. The second row is two, one, one. The third row is one, minus one, two. Okay? Now we expand along the first row. The middle entry, two, sits in the second column. Which sign does its term get, plus or minus? Minus, that is correct, because the signs along the row go plus, minus, plus. So we get one times the bracket two plus one, minus two times the bracket four minus one, plus lambda times the bracket minus two minus one. Oh no, careful here: take plus for this middle term, and lambda comes out wrong. Right? Simplify each bracket, and we get three, then minus six, then minus three lambda. Three minus six is minus three, so the total is minus three, minus three lambda. Clear? Set it equal to zero; the arrow means implies. Minus three, minus three lambda, equals zero, which gives lambda equal to minus one. So at this lambda the box goes flat. That solves our puzzle! Got it?

5:27Example 2: a x (b x c) = b/2, find angles of a with b and c

Now a harder one, problem twenty six, marked medium. Vectors a, b and c are unit vectors, each of length one. Vector a, cross, b cross c in brackets, equals one half of vector b, and b and c are non parallel. Find the angles that vector a makes with b and with c. Pause the video and try it first. Done? Okay, let's check. First we open the vector triple product, and the key question is this: which dot product goes with vector b, a dot c or a dot b? It is a dot c, that is correct! The outer vector is vector a, the far one is c, and the near one is b. So we get a dot c, times b, minus a dot b, times c, in that order. Swap these two dot products, and everything after goes wrong. Okay? Now set this equal to the right hand side, one half of vector b. Right? It is tempting to compare the coefficients on both sides straight away. But tell me, when is that allowed? Only when b and c are non parallel, yes, exactly! If c were parallel to b, it would be a multiple of b, and the terms could mix. Here they are not parallel, so we compare. Clear? For the coefficient of b, the left side has a dot c and the right side has one half, so a dot c is one half. For the coefficient of c, the left side has minus a dot b, and the right side has no c term at all, so a dot b is zero. Got it? Now use the word unit: both lengths are one, so the dot product is just the cosine of the angle between them. With me? So the cosine of the angle with b is zero, which gives ninety degrees; the small circle is the degree sign. And the cosine of the angle with c is one half, which gives sixty degrees. So the non parallel condition did the real work here. Very good!

7:28Quick revision: coplanarity, volume, outer dot far rule

So, quick revision. The scalar triple product is zero exactly for coplanar vectors, and its modulus is the volume of the box. The vector triple product opens as outer dot far, times near, minus outer dot near, times far. And compare the coefficients of b and c only when they are non parallel. Now go and try the practice problems on this page. All the best!