Reciprocal Systems and Vector Equations

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. An unknown vector r has three dot products: r dot a is one, r dot b is two, and r dot c is three. Can you find the vector r itself? With i hat, j hat and k hat it is easy, since those dot products are just the components. But what if a, b and c are slanted? Today we build partner vectors, the reciprocal system, that solve this in one line. It is a handy tool for JEE vector questions. So, let's begin.

0:35Building the reciprocal system

So, we start with three vectors a, b and c; the small arrow on top marks a vector. They must be non-coplanar, not all in one plane, so their scalar triple product, a b c in square brackets, is not zero. The slashed equals sign means not equal, and the double-headed arrow means if and only if. Okay? Now we build a partner for a, called a prime; the small dash at the top right is read as prime. So a prime equals b cross c, over a b c; the cross sign means the cross product. So the top is perpendicular to both b and c, okay? And the bottom is never zero for non-coplanar vectors. With me? The other two follow the cyclic order. So b prime is c cross a, over a b c, and c prime is a cross b, over the same a b c. Oh no, careful here: for b prime it is c cross a, not a cross c, because swapping the order flips the sign. Clear?

1:37Dot products: one with the partner, zero otherwise

Now let me ask you something. What is a dot a prime? Here the dot is the dot product. It is one, yes, exactly! On top we get a dot, b cross c in brackets, and that is the scalar triple product a b c itself. Top and bottom are the same, so it is one. Okay? In the same way, b dot b prime and c dot c prime are also one, by the cyclic property. Right? Now a mixed one: what is a dot b prime? It is zero, because on top we get a dot, c cross a, and c cross a is perpendicular to a, so this is zero. The same happens for every mixed pair, like a dot c prime, or b dot a prime. They all vanish. Got it? One more result to remember. The triple product of the primes, a prime b prime c prime in square brackets, equals one over a b c. With me?

2:33Any vector from three dot products

Now comes the big result of this lesson. Any vector r equals, r dot a in brackets, times a prime, plus r dot b, times b prime, plus r dot c, times c prime. Each dot product goes with its own partner. Okay? Why is this true? If we dot both sides with a, which terms on the right survive? Only the first term, yes! Here a dot a prime is one, while a dot b prime and a dot c prime are zero. So the right side is just r dot a, exactly the left side. The same check works with b and c. Clear? So here comes the payoff. Suppose a question gives r dot a equals p, r dot b equals q, and r dot c equals s, okay? The tempting answer is p times a, plus q times b, plus s times c, right? No, no, wait, that is exactly the trap! The numbers p, q and s go with the primes. The arrow means implies: r equals p times a prime, plus q times b prime, plus s times c prime. And it is the unique solution, because a, b and c are non-coplanar. With me? Now a quick one: if p, q and s are all zero, what is the vector r? It is the zero vector, yes, the zero with an arrow on top! Geometrically, a vector perpendicular to three non-coplanar vectors must be the zero vector. Got it?

4:08Example 1: reciprocal system of i, i + j and i + j + k

Okay, now a course example. Vector a is i hat, b is i hat plus j hat, and c is i hat plus j hat plus k hat; the hats mark unit vectors along the axes. Find the reciprocal system, and check that a dot a prime is one, and a dot b prime is zero. Pause the video and try it first. Done? Okay, let's check. In components, in round brackets, b is one, one, zero, and c is one, one, one. For b cross c, the first component is one times one, minus zero times one, giving one. The second is zero times one, minus one times one, giving minus one. The third is one times one, minus one times one, giving zero. Okay? Now, before we divide by anything, what must we check? Yes, exactly, the scalar triple product must not be zero! It is a dot, b cross c, so one, zero, zero, dot, one, minus one, zero, which gives one. That is not zero, so the reciprocal system exists. Right? Next, c cross a, worked out the same way, comes to zero, one, minus one. And a cross b comes to zero, zero, one. Clear? Now divide each by the triple product, which is one. So a prime is i hat minus j hat, b prime is j hat minus k hat, and c prime is just k hat. With me? Now let us do the check. Vector a dot a prime is one times one, plus zero, plus zero, which is one. And a dot b prime is one times zero, plus zero times one, plus zero times minus one, which is zero. Both match perfectly. Got it?

5:58Example 2: find r from r dot a = 1, r dot b = 2, r dot c = 3

Now, remember our opening puzzle? Let's finish it. With the vectors of example one, find the vector r such that r dot a is one, r dot b is two, and r dot c is three. Here is a hint: use the primes we just found. Pause the video and try it first. Done? Okay, let's check. Here p is one, q is two, and s is three. So r equals one times a prime, plus two times b prime, plus three times c prime. The numbers go with the primes, okay? Now put in the components from example one. The vector a prime is one, minus one, zero; b prime is zero, one, minus one; and c prime is zero, zero, one. Right? Add them component by component. The first is one, plus zero, plus zero, which is one. The second is minus one, plus two, plus zero, which is one. The third is zero, minus two, plus three, which is one. So r is one, one, one, that is, i hat plus j hat plus k hat. Clear? Now, how can we check this answer quickly? Write r as x, y, z in components. Then the first component, x, equals r dot a, which is one; next, x plus y equals r dot b, which is two; and x plus y plus z equals r dot c, which is three. So all three components come out as one. Same answer, and our opening puzzle is solved. With me?

7:33Quick revision: a prime, dot products, r from p, q and s

So, quick revision. First, a prime is b cross c over a b c, with b prime and c prime in cyclic order, and a b c must not be zero. Second, each vector dotted with its partner gives one, mixed dot products are zero, and the primes have triple product one over a b c. Third, if r dot a, r dot b and r dot c are p, q and s, then r is p times a prime, plus q times b prime, plus s times c prime. You have just solved a vector equation in one line, well done! Now try the practice problem on this page. All the best!