Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle to start. Two planes meet in a line, and through that line pass infinitely many planes, like the pages of an open book. One neat equation gives them: P one, plus lambda times P two, equals zero. Does it give every one of those planes? Most students say yes, but one plane is missing! Hold that thought. Today we do six quick concept checks from Part C of our three dimensional geometry problem bank, statements JEE can test. So, let's begin.

0:34Two lines perpendicular to a third

So, let us start with lines. Take two lines, both perpendicular to a third line; the upside-down T sign means perpendicular. In one plane, like a page of your notebook, those two lines must be parallel, right? But now we are in space, in three dimensions. Must the two lines still be parallel? Not necessarily, and the axes show it. In the figure, the x, y and z axes meet at the origin, O, at right angles; leave the blue line and its angles for Example one. The x-axis is perpendicular to the z-axis, and so is the y-axis. With me? But the x and y axes are not parallel at all; they meet at the origin, O, at a right angle. Clear? So the answer is yes: in space, two lines can both be perpendicular to a third line without being parallel. This idea comes up often in statement questions. Got it?

1:33The skew-line formula for parallel lines?

Now, distances between lines. On the right are two skew lines: not parallel, and never meeting. L one passes through a one along b one, and L two through a two along b two; lambda and mu, Greek letters, are their parameters. The little arrows mean vectors, and the small numbers below are labels. P Q meets both lines at right angles; its length is the shortest distance, d, and it is parallel to b one cross b two, the two short vertical lines meaning parallel. The formula: d equals the modulus of, a two minus a one, dot, b one cross b two, over the modulus of b one cross b two; the vertical bars mean modulus. Okay? Now, a concept check for you. Can we use this same formula for two parallel lines? We cannot, and here is why. Parallel lines have parallel directions, so b one cross b two is the zero vector; the arrow on the zero says it is a vector. Right? So the bottom is zero, and the top, a dot product with the zero vector, is zero too. Zero over zero gives no answer at all. Clear? For parallel lines, with the same direction b, the course uses a different formula: d equals the modulus of, a two minus a one, cross b, over the modulus of b. Notice that it uses cross here, not dot. With me? It is simply the distance of the point a two, on the second line, from the first line. Got it?

3:06A line and a plane

Next, a line and a plane. The line is r vector equals a vector, plus t times b vector, so b vector is its direction. The plane is n vector dot r vector equals d, and n vector is its normal, the direction perpendicular to the plane. Okay? Now, the angle between this line and the plane. Is it just the angle between the line and the normal? No, and this is a classic trap, because the normal stands at ninety degrees to the plane. So phi, the Greek letter for our angle, is ninety degrees, the small circle meaning degrees, minus the angle with the normal. The two angles are complementary. Right? So the cosine with the normal becomes a sine with the plane: sine phi equals the modulus of b dot n, over the modulus of b, times the modulus of n. Sine, not cosine, so write that down. Clear? Now take a line parallel to the plane. Its direction is perpendicular to the normal, so b dot n is zero. Is every point of such a line at the same distance from the plane? Yes, it is, and here is the reason. Put any point of the line, a plus t times b, into n dot r. We get n dot a, plus t times n dot b, and n dot b is zero, so it is just n dot a, for every value of t. Got it? The distance from the plane depends only on this value, so every point of the line is equally far. That common distance is the distance between the line and the plane. With me?

4:43Example 1: can 1/2, 1/2, 1/2 be direction cosines?

Now Example one. Can one half, one half, one half, in round brackets, be the direction cosines of a line? Pause the video and try it first. Done? Okay, let's check. In the figure, the blue line O P makes angles alpha, beta and gamma, Greek letters, with the axes; its direction cosines l, m and n are their cosines. They must satisfy l squared plus m squared plus n squared equals one; the small two up top means squared. So, square each one half and add. What do you get? Yes, exactly, one quarter three times gives three quarters, and three quarters is not equal to one; the slashed equals sign means not equal. Okay? So the answer is no, these cannot be direction cosines. But they can be direction ratios, because ratios need no such condition. Right? To get the direction cosines, divide each by the square root of three quarters, the sum of the squares; that is root three over two. Clear? One half, divided by root three over two, gives one over root three. So all three direction cosines are one over root three, and their squares, one third each, add up to one. Got it?

6:03Example 2: does P1 + lambda P2 = 0 give every plane?

Now Example two, our puzzle from the start. Does P one, plus lambda times P two, equals zero, give every plane through the line of intersection of P one equals zero and P two equals zero? Pause the video and try it first. Done? Okay, let's check. In the figure, the dark blue planes, P one equals zero and P two equals zero, meet in the black line. The orange plane, P one plus lambda times P two equals zero, and the grey plane are members of this family. Each real lambda gives one plane through the line, and lambda equal to zero gives P one equals zero itself. Okay? Now tell me, which value of lambda gives the second plane, P two equals zero? Let us work it out together. For lambda not zero, divide by lambda: one over lambda, times P one, plus P two, equals zero. Right? To get P two equals zero alone, one over lambda must be zero. No real lambda does that; we would need, in quotes, lambda equal to infinity, the sideways eight, which is not a real number. With me? So here is the answer: the family gives every plane through the line, except P two equals zero itself. That is the missing plane from our puzzle! Clear? And here is a topper's habit: whenever the answer might be P two equals zero itself, test that plane separately. Got it?

7:34Quick revision: the six 3D geometry checks recapped

So, quick revision. In space, two lines perpendicular to a third need not be parallel. For parallel lines, b one cross b two is the zero vector, so use the parallel-line formula. For a line and a plane, use sine; a parallel line keeps one common distance. Direction cosines satisfy l squared plus m squared plus n squared equals one; direction ratios need not. And P one, plus lambda times P two, equals zero, gives every plane through the line except P two equals zero. Six checks, very well done! Now try the practice problem on this page. All the best!