Overview

Transcript

0:00Introduction

Hello everyone! Here is one small puzzle before we start. Take two lines, both perpendicular to a third line. Must those two lines be parallel to each other? On a flat page of your notebook, yes, they must, but what about lines in space? Hold on to your answer for now. This chapter is the problem bank for three dimensional geometry, and that puzzle is one of its concept checks. So let us see how to use it, and solve two of its problems together. Let's begin!

0:30How to use this chapter

So, let's open this chapter. Part A has eight solved problems at JEE Main level: six marked easy, and two marked medium. Okay? Now, how should you use them? The course says: attempt Part A as a thirty minute test, after Chapters one to four; the short dash means to. Right? So, a quick question for you. Eight problems in thirty minutes. How much time can you give each problem? Thirty over eight gives three point seven five. So just under four minutes a problem, with a clock running. Clear? Then Part B: six longer problems at JEE Advanced level, five of them marked hard, and one marked medium. With me? And Part C: six concept checks. They test ideas often asked in statement form: you decide whether a claim is true, and say why. Got it?

1:27Part C: a statement needs a reason

Now, two statements from Part C, the concept checks. In the figure, a line O P makes angles alpha, beta and gamma, three Greek letters, with the axes. Their cosines, l, m and n, are the direction cosines, and they satisfy l squared plus m squared plus n squared equals one. So, the first check: can one half, one half, one half be the direction cosines of a line? Let us square them. One quarter plus one quarter plus one quarter is three quarters, which is not equal to one; the slashed equals sign means not equal. So no, they cannot be direction cosines. Right? But they are fine as direction ratios, which need not satisfy that condition. For the direction cosines, the d c's, divide each by root three over two, the root of the sum of the squares. Each becomes one over root three; the tick shaped sign is the square root. Clear? Now the second check, which is our puzzle. The upside down T on screen means perpendicular to. In space, can two lines both be perpendicular to a third line, without being parallel to each other? Look at the axes in the figure. The x axis and the y axis are both perpendicular to the z axis, and yet they are not parallel; they meet at right angles. So yes, it can happen, and that solves our puzzle: the flat page rule does not carry over to space. In Part C, the trap is a quick answer from habit, so always give a reason. With me?

3:02Example 1: shortest distance between two skew lines

Now, Example one, from Part A, marked medium. Find the shortest distance between the lines x over one equals y over one equals z over one, and x equals one, y over one equals z over minus one. Remember, under four minutes. Pause the video and try it first. Done? Okay, let's check. The first line, L one, passes through the origin, zero, zero, zero; in the figure, that point is vector a one. Its direction is one, one, one, and we call it vector b one; the arrow on top marks a vector, and the small one below is a subscript. Okay? Now the second line. Many students see x equals one, and write its direction ratios as one, one, minus one, right? No, no, wait, that is the trap! Here x stays fixed at one along the whole line, so its direction ratio along x is zero. So L two passes through one, zero, zero, which is vector a two, with direction vector b two equal to zero, one, minus one. Clear? In the figure, lambda and mu, two Greek letters, slide along the two skew lines. The shortest distance is along the common perpendicular P Q, parallel to b one cross b two; the two small vertical lines mean parallel. Here, b one cross b two, with the small cross sign, is minus two, one, one. Not the zero vector, so the lines are not parallel, and the skew formula applies. With me? Now the numerator. Vector a two minus vector a one is one, zero, zero. Its dot product with minus two, one, one is minus two; the small dot means the dot product. This is the scalar triple product. Right? So d, the distance, is the modulus of minus two, over the modulus of b one cross b two; the vertical bars mean the modulus. The root of four plus one plus one is root six. So d equals two over root six, which is the root of two thirds. Not zero, so these lines never meet: they really are skew. Got it?

5:18Example 2: prove two lines coplanar and find their plane

Now a harder one, from Part B, marked hard. Show that x plus three over minus three, equals y minus one over one, equals z minus five over five, and x plus one over minus one, equals y minus two over two, equals z minus five over five, are coplanar, which means they lie in one plane. Then find that plane. Pause the video and try it first. Done? Okay, let's check. First, a point on the first line. We see x plus three, y minus one, z minus five, so the point is three, minus one, minus five, right? Oh no, careful here, because the signs flip! Here x plus three means x minus, minus three, so the point, vector a one, is minus three, one, five. And the denominators give the direction, vector b one: minus three, one, five again. Same way, vector a two and vector b two are both minus one, two, five. Okay? Now, vector a two minus vector a one. Minus one plus three is two, two minus one is one, and five minus five is zero. So we get two, one, zero. Right? Next, b one cross b two. The first component is five minus ten, which is minus five. The second is minus five plus fifteen, which is ten. And the third is minus six plus one, which is minus five. Clear? Now the dot product. Two times minus five is minus ten, one times ten is ten, and zero times minus five is zero. The total is zero. So what does a zero triple product tell us about these lines? Yes, exactly, the lines are coplanar! And b one cross b two is not zero, so they are not parallel, and the plane's normal is along it, perpendicular to both directions. Divide by minus five, and the normal is parallel to one, minus two, one. Through the point minus three, one, five, the plane is x plus three, minus two times y minus one, plus z minus five, equals zero. With me? Expand it: three plus two minus five is zero, so the plane is x minus two y plus z equals zero. Now check the second line's point, minus one, two, five: minus one, minus four, plus five gives zero. And its direction, b two, is also minus one, two, five, so the same sum shows it is perpendicular to the normal. Its point lies in the plane and its direction is perpendicular to the normal, so the whole second line lies in it too. Got it?

8:04Quick revision: direction cosines, skew vs coplanar test

So, quick revision. Part A is a thirty minute test, Part B is at JEE Advanced level, and in Part C you give reasons. Direction cosines must satisfy l squared plus m squared plus n squared equals one; direction ratios need not. The scalar triple product: not zero means skew, zero means coplanar, and for parallel lines, the distance needs its own formula. Now open Part A, set a thirty minute clock, and start solving. All the best!