Solving Vector Equations

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. If three times x equals twelve, you divide by three, and x is four, right? Now try this: a cross x equals b, where a and b are known vectors. Can you divide by vector a? Of course not, there is no division by a vector! And such an equation can have no solution at all, or infinitely many. Today we learn the standard moves that solve it, and they come up often in JEE vector questions. So, let's begin.

0:34Move 1: bring everything to one side

So, here is move one: r cross a equals b cross a. The small arrow on top marks a vector, and the cross sign is the cross product. The slashed equals sign means vector a is not the zero vector, the zero with an arrow. Now tell me, can we just 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. Okay? Remember, a cross product with a non-zero vector is zero only when the other vector is zero, or parallel to it. Either way, r minus b equals t times a, for some number t. The double arrow means implies, so r equals b plus t times a. Not one answer, but a whole line of answers! Clear? So one unknown number, t, is left. To fix t, the question must give one more condition, a scalar one, like a dot product equal to a number. Got it?

1:41Move 2: when does a cross x = b have a solution?

Now let us look at move two. This time the unknown, x, sits inside the cross product: a cross x equals b, with vector a non-zero again. Let me ask you this. Whatever x is, a cross x is always perpendicular to one known vector. Which one? It is a, yes, exactly! So dot both sides with a; the dot in the middle is the dot product. We get a dot, a cross x in brackets, equals a dot b. And on the left, a dot, a cross x, is zero, because a cross x is perpendicular to a. Look, the unknown has disappeared! Right? So the right side must be zero too. The equation a cross x equals b has a solution only if a dot b is zero. Oh no, careful here: if a dot b is not zero, there is no x at all, so do not start calculating! With me?

2:36Move 3: cross with a to find x

Now suppose a dot b is zero. To pin x down, we need one more condition: a dot x equals c, where c is a number. But the vector x is stuck inside a cross product. How do we pull it out? We cross both sides with the known vector a, yes! So a cross, a cross x in brackets, equals a cross b. Okay? Now expand the left side with a special case of the vector triple product, from Chapter four. It gives a dot x, times a, minus mod a squared, times x. The vertical bars give the modulus, the length of vector a, and the small two on top means squared. Clear? But a dot x is just c! So c times a, minus mod a squared times x, equals a cross b. Look, x is free now. Right? Now solve for x. Move a cross b to the other side, and minus a cross b is the same as plus b cross a, because swapping the order flips the sign. Then divide by mod a squared, which is allowed because a is non-zero. So x equals c times a, plus b cross a, all over mod a squared. Oh no, careful here, it is b cross a on top, not a cross b. Got it?

3:54Topper's trick: dot or cross?

So here is the topper's trick, in two lines. If you want a number, like a dot x, dot the equation with a vector that kills the unknown terms, as in move two. With me? And to remove a cross product, cross with a known vector and expand, as in move three. Okay?

4:11Example 1: r cross a = b cross a with r dot (i+2j+k) = 4

Okay, now a course example. Vector a is i hat plus j hat plus k hat, and vector b is two i hat minus j 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 plus k hat, equals four. Pause the video and try it first. Done? Okay, let's check. First, the tempting shortcut. Cancel a, and r equals b, right? Let's test it. Vector b dot, i hat plus two j hat plus k hat, is two, minus two, plus zero, which is zero. But we need four! So r equals b is wrong. Clear? Now do it properly: by move one, r equals b plus t times a. In components, in round brackets, b is two, minus one, zero, and a is one, one, one. So r is two plus t, then minus one plus t, then t. Okay? Now the dot condition. Two plus t, plus two times, minus one plus t, plus t. The constants cancel, leaving four times t. Four times t equals four, so t equals one. With me? Put t equal to one, and r is three, zero, one. So r equals three i hat plus k hat. Right? And now let us do a quick check. Here, r minus b is one, one, one, which is exactly a, and a cross a is the zero vector. So it works. Perfect, got it?

5:46Example 2: solve a cross x = b with a dot x = 3

Now a harder one. Solve a cross x equals b, and a dot x equals three, where a is i hat minus j hat plus k hat, and b is i hat plus j hat. Here is a hint: use moves two and three. Pause the video and try it first. Done? Okay, let's check. Before using any formula, what must we check first? Yes, exactly, we check a dot b! It is one minus one plus zero, which is zero. So a solution exists. Right? Next, the cross product, with the component formula. The first component is minus one times zero, minus one times one, giving minus one. The second is one times one, minus one times zero, giving one. The third is one times one, minus, minus one times one, giving two. So a cross b is minus one, one, two. But the formula needs b cross a, so flip every sign, and b cross a is one, minus one, minus two. Clear? Here, c is three, and mod a squared is one plus one plus one, which is three. Okay? Now plug in. So x is one third of, three times, one, minus one, one, plus, one, minus one, minus two. Three plus one is four, minus three minus one is minus four, and three minus two is one. So x is one third of, four, minus four, one. With me? That is, x equals one third of, four i hat minus four j hat plus k hat, the course answer. Got it? And we should always check our answer. Here, a cross x comes out as one, one, zero, which is b, and a dot x is four plus four plus one, over three, which is three. Both conditions hold, and our opening puzzle is solved without dividing by a vector!

7:35Quick revision: r = b + ta, formula for x, dot or cross

So, let us do a quick revision. If r cross a equals b cross a, then r equals b plus t times a, and a scalar condition fixes t. Next, a cross x equals b needs a dot b to be zero, and with a dot x equal to c, x is c times a plus b cross a, over mod a squared. And dot to find a number, cross to remove a cross product. Now go and try the practice problem on this page. All the best!