Collinearity, Coplanarity and Linear Combinations

Transcript

0:00Introduction

Hello everyone! Let me start with a quick question. Suppose x times vector a, plus y times vector b, is the zero vector. Does that force x and y to be zero? Most students say yes, straight away. But what if b is just two times a, for example? Then two times a, minus b, is the zero vector, and those numbers are not zero at all! Today we find out exactly when we may compare coefficients, and how to test whether points lie on one line or in one plane. This idea comes up often in JEE vectors, so let's begin.

0:39Linear combinations and independence

So, first the words. The little arrow on top means a vector, and x, y and z are ordinary numbers, scalars. Vector r equals x times a, plus y times b, plus z times c, and we call r a linear combination of a, b and c, okay? Now, the zero with an arrow is the zero vector, and the arrow pointing right means implies. The vectors are linearly independent if x times a, plus y times b, plus z times c, can be the zero vector only when x, y and z are all zero. Otherwise they are dependent. Right? Now take just two vectors. The arrow pointing both ways means if and only if. Two vectors are independent exactly when they are non collinear, not along the same or parallel lines. That was our opening puzzle: when b is two times a, the two are collinear, so they are dependent. Clear? Now let us take three vectors. Suppose no two of them are parallel. Are the three vectors then independent? Not necessarily, and this is a common trap! Three vectors can point in different directions and still lie in one plane, like three arrows drawn on your notebook page. So three vectors are independent exactly when they are non coplanar. Remember, any two vectors are coplanar, but three need not be. Got it?

2:04Unique form: why we may compare coefficients

Now, here is why independence matters so much. Take two vectors a and b that are non collinear. Every vector r in their plane can be written as x times a, plus y times b, and uniquely: only one pair of numbers x and y does the job. Okay? In space, we need three. If a, b and c are non coplanar, every vector in space is x times a, plus y times b, plus z times c, and again, uniquely. With me? So let me ask you this: if two such expressions give the same vector, what can you say about their numbers? Yes, exactly, the numbers must match, one by one! So we may compare coefficients: the number in front of each vector must be the same on both sides. But only along independent vectors, never along dependent ones. Right?

2:57Three points on a line, four in a plane

Now let us move from vectors to points. Let the points A, B and C have position vectors a, b and c, in that order. The arrow over A B means the vector from A to B, okay? The three points are collinear, on one straight line, exactly when vector A B is lambda times vector A C, where lambda, the Greek letter, is a scalar. The vectors are parallel and share the point A, so the points sit on one line. Clear? There is a second test, straight from the position vectors. Suppose x times a, plus y times b, plus z times c, is the zero vector, with x, y and z not all zero. Is that enough for the points to be collinear? No, no, careful, that alone is not enough! The coefficients must also add up to zero: x plus y plus z equals zero. With both conditions, the points are collinear. So here is a topper's habit: when you see a relation between position vectors, first add up the coefficients. Got it? Now the same idea, one level up. Four points A, B, C and D, with position vectors a, b, c and d, are coplanar, in one plane, exactly when x times a, plus y times b, plus z times c, plus w times d, is the zero vector, okay? And again, with the four coefficients adding up to zero, and not all of them zero. So three points with a zero sum lie on a line, and four points with a zero sum lie in a plane. With me?

4:31Example 1: find lambda so three points are collinear

Now, example one. Here i hat and j hat, with the little cap on top, are the unit vectors along the x and y axes. The position vectors are sixty i hat plus three j hat, forty i hat minus eight j hat, and lambda i hat minus fifty two j hat. For what value of lambda are the three points collinear? Pause the video and try it first. Done? Okay, let's check. Call the points A, B and C, in that order. Vector A B is B minus A, terminal minus initial, and the brackets hold its i and j components. Forty minus sixty, and minus eight minus three, give minus twenty and minus eleven, okay? In the same way, vector A C has components lambda minus sixty, and minus fifty two minus three, which is minus fifty five. Right? For collinear points, vector A C must be some multiple of vector A B, and we must find it. Now tell me, which component should you use to find that multiplier? The j component, of course, because there both numbers are known! Minus fifty five is five times minus eleven, so vector A C must equal five times vector A B, as the line shows. Clear? Now the i components must match too. Lambda minus sixty equals five times minus twenty, which is minus one hundred. So lambda equals minus forty. Careful with the sign here: minus one hundred plus sixty gives minus forty, not minus one hundred and sixty. Got it?

6:03Example 2: find lambda so three vectors are coplanar

Now a harder one. The vectors a, b and c are non coplanar. Find lambda so that these three vectors are coplanar: a, minus two times b, plus three times c; then minus two times a, plus three times b, minus four times c; and a, minus three times b, plus lambda times c, okay? Pause the video and try it first. Done? Okay, let's check. The first two vectors have coefficients one, minus two, three, and minus two, three, minus four. These are not in the same ratio, so the two are not parallel, and they are independent. With me? Now, all three must be coplanar, and the first two already fix that plane. So what must the third vector be? Yes, exactly, a combination of the first two! So a, minus three times b, plus lambda times c, equals x times the first vector, plus y times the second. Okay? Now we compare coefficients, allowed only because a, b and c are non coplanar, as given. The coefficient of vector a gives x minus two times y equals one. The coefficient of vector b gives minus two times x, plus three times y, equals minus three. Right? Solving the two together gives y equals one, and x equals three. Put them back in, and both equations check out. Clear? And now the coefficient of vector c, which gives lambda. Lambda equals three times x, minus four times y, that is nine minus four, so lambda is five. See the plan: two coefficients find x and y, and the third coefficient finds lambda, and that is the whole method!

7:47Quick revision: independence, collinear and coplanar tests

So, quick revision. Two vectors are independent when they are non collinear, and three when they are non coplanar. Compare coefficients only along independent vectors. And when a combination of position vectors is the zero vector, with coefficients adding to zero, not all zero, three points are collinear, and four are coplanar. Now go and try the practice problems on this page. All the best!